Does this limit exist on $mathbb R^2$
$begingroup$
$(x,y) in mathbb R^2$
$$lim_{(x,y)to(1,1)} frac{(x-y)^{(x-y)}} {(x-y)}$$
Does the limit above exist? Neither I could compute it nor I could find directions which have different limit values. Can someone help me please? If this limit exists how can I compute it if doesn't exist which directions should I use?
Thanks a lot in advance
calculus analysis multivariable-calculus
asked Dec 7 '18 at 11:15
user519955user519955
312111
$endgroup$
add a comment |
$begingroup$
$(x,y) in mathbb R^2$
$$lim_{(x,y)to(1,1)} frac{(x-y)^{(x-y)}} {(x-y)}$$
Does the limit above exist? Neither I could compute it nor I could find directions which have different limit values. Can someone help me please? If this limit exists how can I compute it if doesn't exist which directions should I use?
Thanks a lot in advance
calculus analysis multivariable-calculus
asked Dec 7 '18 at 11:15
user519955user519955
312111
$endgroup$
$begingroup$
You need to consider paths within the domain of definition for the function that is $$D={(x,y)inmathbb{R^2}:x-y>0}$$
$endgroup$
– gimusi
Dec 7 '18 at 12:32
2
$begingroup$
Some sources would insist that for the limit to exist the function should be defined in a punctured neighborhood of $(1,1)$. Other sources would never worry about points outside the domain of definition, in which case you are restricted to the half-plane $x>y$. Check what is the definition in use in your book, please!
$endgroup$
– Jyrki Lahtonen
Dec 7 '18 at 13:16
1
$begingroup$
@JyrkiLahtonen We have already discussed that point recently. Not all sources are to be considered at the same level. Using the first definition is reasonable to a lower (high school) level whereas in a more advanced context the second definition (according to Rudin) should be adopted. Notably according to the first definiiton the interesting question on the present limit would become a trivial question on the domain of the function.
$endgroup$
– gimusi
Dec 7 '18 at 14:32
$begingroup$
Indeed, @gimusi. I do remember that discussion. I still think that it is important for the askers to be aware of this (and to explain). What you say about the question becoming trivial otherwise is, of course, valid.
$endgroup$
– Jyrki Lahtonen
Dec 7 '18 at 17:28
add a comment |
$begingroup$
$(x,y) in mathbb R^2$
$$lim_{(x,y)to(1,1)} frac{(x-y)^{(x-y)}} {(x-y)}$$
Does the limit above exist? Neither I could compute it nor I could find directions which have different limit values. Can someone help me please? If this limit exists how can I compute it if doesn't exist which directions should I use?
Thanks a lot in advance
calculus analysis multivariable-calculus
asked Dec 7 '18 at 11:15
user519955user519955
312111
$endgroup$
$(x,y) in mathbb R^2$
$$lim_{(x,y)to(1,1)} frac{(x-y)^{(x-y)}} {(x-y)}$$
Does the limit above exist? Neither I could compute it nor I could find directions which have different limit values. Can someone help me please? If this limit exists how can I compute it if doesn't exist which directions should I use?
Thanks a lot in advance
calculus analysis multivariable-calculus
calculus analysis multivariable-calculus
asked Dec 7 '18 at 11:15
user519955user519955
312111
asked Dec 7 '18 at 11:15
user519955user519955
312111
asked Dec 7 '18 at 11:15
user519955user519955
312111
asked Dec 7 '18 at 11:15
user519955user519955
312111
asked Dec 7 '18 at 11:15
user519955user519955
312111
312111
$begingroup$
You need to consider paths within the domain of definition for the function that is $$D={(x,y)inmathbb{R^2}:x-y>0}$$
$endgroup$
– gimusi
Dec 7 '18 at 12:32
2
$begingroup$
Some sources would insist that for the limit to exist the function should be defined in a punctured neighborhood of $(1,1)$. Other sources would never worry about points outside the domain of definition, in which case you are restricted to the half-plane $x>y$. Check what is the definition in use in your book, please!
$endgroup$
– Jyrki Lahtonen
Dec 7 '18 at 13:16
1
$begingroup$
@JyrkiLahtonen We have already discussed that point recently. Not all sources are to be considered at the same level. Using the first definition is reasonable to a lower (high school) level whereas in a more advanced context the second definition (according to Rudin) should be adopted. Notably according to the first definiiton the interesting question on the present limit would become a trivial question on the domain of the function.
$endgroup$
– gimusi
Dec 7 '18 at 14:32
$begingroup$
Indeed, @gimusi. I do remember that discussion. I still think that it is important for the askers to be aware of this (and to explain). What you say about the question becoming trivial otherwise is, of course, valid.
$endgroup$
– Jyrki Lahtonen
Dec 7 '18 at 17:28
add a comment |
$begingroup$
You need to consider paths within the domain of definition for the function that is $$D={(x,y)inmathbb{R^2}:x-y>0}$$
$endgroup$
– gimusi
Dec 7 '18 at 12:32
2
$begingroup$
Some sources would insist that for the limit to exist the function should be defined in a punctured neighborhood of $(1,1)$. Other sources would never worry about points outside the domain of definition, in which case you are restricted to the half-plane $x>y$. Check what is the definition in use in your book, please!
$endgroup$
– Jyrki Lahtonen
Dec 7 '18 at 13:16
1
$begingroup$
@JyrkiLahtonen We have already discussed that point recently. Not all sources are to be considered at the same level. Using the first definition is reasonable to a lower (high school) level whereas in a more advanced context the second definition (according to Rudin) should be adopted. Notably according to the first definiiton the interesting question on the present limit would become a trivial question on the domain of the function.
$endgroup$
– gimusi
Dec 7 '18 at 14:32
$begingroup$
Indeed, @gimusi. I do remember that discussion. I still think that it is important for the askers to be aware of this (and to explain). What you say about the question becoming trivial otherwise is, of course, valid.
$endgroup$
– Jyrki Lahtonen
Dec 7 '18 at 17:28
$begingroup$
You need to consider paths within the domain of definition for the function that is $$D={(x,y)inmathbb{R^2}:x-y>0}$$
$endgroup$
– gimusi
Dec 7 '18 at 12:32
$begingroup$
You need to consider paths within the domain of definition for the function that is $$D={(x,y)inmathbb{R^2}:x-y>0}$$
$endgroup$
– gimusi
Dec 7 '18 at 12:32
2
2
$begingroup$
Some sources would insist that for the limit to exist the function should be defined in a punctured neighborhood of $(1,1)$. Other sources would never worry about points outside the domain of definition, in which case you are restricted to the half-plane $x>y$. Check what is the definition in use in your book, please!
$endgroup$
– Jyrki Lahtonen
Dec 7 '18 at 13:16
$begingroup$
Some sources would insist that for the limit to exist the function should be defined in a punctured neighborhood of $(1,1)$. Other sources would never worry about points outside the domain of definition, in which case you are restricted to the half-plane $x>y$. Check what is the definition in use in your book, please!
$endgroup$
– Jyrki Lahtonen
Dec 7 '18 at 13:16
1
1
$begingroup$
@JyrkiLahtonen We have already discussed that point recently. Not all sources are to be considered at the same level. Using the first definition is reasonable to a lower (high school) level whereas in a more advanced context the second definition (according to Rudin) should be adopted. Notably according to the first definiiton the interesting question on the present limit would become a trivial question on the domain of the function.
$endgroup$
– gimusi
Dec 7 '18 at 14:32
$begingroup$
@JyrkiLahtonen We have already discussed that point recently. Not all sources are to be considered at the same level. Using the first definition is reasonable to a lower (high school) level whereas in a more advanced context the second definition (according to Rudin) should be adopted. Notably according to the first definiiton the interesting question on the present limit would become a trivial question on the domain of the function.
$endgroup$
– gimusi
Dec 7 '18 at 14:32
$begingroup$
Indeed, @gimusi. I do remember that discussion. I still think that it is important for the askers to be aware of this (and to explain). What you say about the question becoming trivial otherwise is, of course, valid.
$endgroup$
– Jyrki Lahtonen
Dec 7 '18 at 17:28
$begingroup$
Indeed, @gimusi. I do remember that discussion. I still think that it is important for the askers to be aware of this (and to explain). What you say about the question becoming trivial otherwise is, of course, valid.
$endgroup$
– Jyrki Lahtonen
Dec 7 '18 at 17:28
add a comment |
4 Answers
4
active
oldest
votes
$begingroup$
Hint: $displaystylelim_{xto0}x^x=1$
answered Dec 7 '18 at 11:24
José Carlos SantosJosé Carlos Santos
166k22132235
$endgroup$
$begingroup$
If I use it it seems like the limit is infinity. What is my mistake?
$endgroup$
– user519955
Dec 7 '18 at 12:02
$begingroup$
$lim_{(x,y)to(1,1)} frac{(x-y)^{(x-y)}} {(x-y)}$ = $lim_{(x,y)to(1,1)} (x-y)^{(x-y)}. lim_{(x,y)to(1,1)} frac{1} {(x-y)}$ $to$ $1. infty$
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– user519955
Dec 7 '18 at 12:04
2
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I didn't write them.
$endgroup$
– José Carlos Santos
Dec 7 '18 at 12:11
1
$begingroup$
@JoséCarlosSantos Divergent limits are not said to exist. A limit exists if it is real, finite and unique. Unless otherwise stated, existence of a limit is checked in the real numbers, not the extended real line.
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– Shubham Johri
Dec 7 '18 at 12:22
1
$begingroup$
So, for you the limit $lim_{xto+infty}x$ doesn't exist. I find it much more natural to say that it is $+infty$.
$endgroup$
– José Carlos Santos
Dec 7 '18 at 12:26
|
show 4 more comments
$begingroup$
No, this limit doesn't exist. Chose the path $y=x$. Alternatively, let $x-y=m$. Then the question is just about evaluating $lim_{mto0} frac{m^m}m=lim_{mto0}m^{m-1}$ which doesn't exist.
answered Dec 7 '18 at 11:29
Shubham JohriShubham Johri
5,204718
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Come on, you can't choose the path $y=x$...
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– dallonsi
Dec 7 '18 at 11:51
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Why can't we chose $y=x$? It is pretty much a part of the domain.
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– Shubham Johri
Dec 7 '18 at 11:53
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if $y=x$ then you make a division by $0$ ! No matter what your numerator is, you still write a division by zero.
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– dallonsi
Dec 7 '18 at 11:54
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That is actually the point.
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– Shubham Johri
Dec 7 '18 at 11:55
1
$begingroup$
@gimusi $infty$ is not an object, it's a concept. A limit exists if it is equal to some number (not $infty$). Do your research online.
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– Algebear
Dec 7 '18 at 12:24
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show 15 more comments
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This limit does not exist.
Let $varepsilon in mathbb{R}^*_+$.
Take $x=1 + varepsilon$ and $y= 1$.
Then rewrite the quotient as :
$$varepsilon^{varepsilon -1} = e^{(varepsilon -1) ln (varepsilon)}.$$
Since $lim_{varepsilon to 0^+} varepsilon ln (varepsilon) =0$ and $lim_{varepsilon to 0^+} - ln (varepsilon) = +infty$, then the limit of $frac{(x-y)^{(x-y)}} {(x-y)}$ is $+infty$.
Now you want to apply the same trick with $-varepsilon$ to get a different limit. The thing is that the power of a negative number might be not be a real number (take for example the square root of $-2$). You can get rid off this problem by using complexe numbers.
Thus, take now $x = 1 - varepsilon$ and $y=1$. The quotient is now:
$$ frac{(-varepsilon)^{-varepsilon}}{-varepsilon} = e^{-ifrac{pi}{2}varepsilon}frac{varepsilon^{-varepsilon}}{-varepsilon} = e^{-ifrac{pi}{2}varepsilon}frac{1}{-varepsilon^{varepsilon + 1}} $$
from here you see that $e^{-ifrac{pi}{2}varepsilon}$ tends to 1 and the denominator $-varepsilon^{varepsilon + 1} $ goes to $0^{-}$. Thus the limit is now $-infty$.
Since you can find two "paths" $(x,y)$ which goes to $1$ and which give different limits, the limit $lim_{(x,y)to(1,1)} frac{(x-y)^{(x-y)}} {(x-y)}$ does not exist.
answered Dec 7 '18 at 11:49
dallonsidallonsi
1187
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I think that your conclusion is wrong, the limit exists and it is $+infty$.
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– gimusi
Dec 7 '18 at 12:13
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Huw do you define $(-varepsilon)^{-varepsilon}$?
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– gimusi
Dec 7 '18 at 12:38
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well if $x,y in mathbb{R}$, you can always say that $x^y =|x|^y times e^{iarg(x) times y}$.
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– dallonsi
Dec 7 '18 at 13:17
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another way to say it is : $lim_{(x,y) to (1,1)} (x-y)^{(x-y)} = 1$, right ? So now, if you divide it by $(x-y)$, you can find a path that leads the quotient to $+infty$ (if $(x-y) to 0^+$) or $-infty$ (if $(x-y) to 0^{-}$). @gimusi, do you agree ?
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– dallonsi
Dec 7 '18 at 13:20
add a comment |
$begingroup$
We have that by $x-y =t to 0^+$ the given limit reduces to
$$lim_{(x,y)to(1,1)} frac{(x-y)^{(x-y)}} {(x-y)}=lim_{tto 0^+} frac{t^t} {t}to infty$$
indeed the points $(x-y)<0$ are not included within the domain of definition for the function that is $$D={(x,y)inmathbb{R^2}:x-y>0}$$
Refer also to the related
- domain of $x^x$
Edit for a remark
As noticed by Jyrky Lahtonen in the comment "some sources would insist that for the limit to exist the function should be defined in a punctured neighborhood of $(1,1)$. Other sources would never worry about points outside the domain of definition, in which case you are restricted to the half-plane $x>y$".
That important issue has been deeply discussed here
- What is $lim_{x to 0}frac{sin(frac 1x)}{sin (frac 1 x)}$ ? Does it exist?
My observation is that not all sources have to be considered at the same level. Using the first definition is a reasonable way to deal with limits at a lower (high school) level whereas, in a more advanced context, the second more general definition (according to Rudin) should be adopted.
Moreover, note that, according to the first definition, the discussion of limits would reduce, as in the present case, to a (more or less) trivial determination on the domain of definition for functions which is of course a very different topic.
Therefore, when we deal with limits, I strongly suggest to refer to the more general definition excluding points outside the domain of definition for the function considered.
answered Dec 7 '18 at 12:12
gimusigimusi
93k84494
$endgroup$
1
$begingroup$
Why aren't $(x-y) lt 0$ included in the domain? Domain is all $mathbb R^2$
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– user519955
Dec 7 '18 at 12:17
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@user519955 The key point is exactly that one. For which values is $f(x)=x^x$ defined?
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– gimusi
Dec 7 '18 at 12:18
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@user519955 I suggest to refer to that OP and notably to the answer HERE. It often a point of counfusion. Do not hesitate to ask for any clarifiaction on that.
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– gimusi
Dec 7 '18 at 12:24
1
$begingroup$
I will glance it. But domain seems like $mathbb R - {0} cup {1/(2n | nin mathbb N}$ for $x^x$ not $x>0$ to me. For example $(-2)^{(-2)}=1/4$
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– user519955
Dec 7 '18 at 12:48
1
$begingroup$
@user519955 That’s a good point, I agree that the expression $x^x$ is defined for some values as also for example $(-1/3)^{-1/3}$ but in the calculus context the domain for the function $f(x)=x^x$ is defined only for $x>0$. Refer to that OP for a discussion on that.
$endgroup$
– gimusi
Dec 7 '18 at 13:00
add a comment |
Your Answer
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4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Hint: $displaystylelim_{xto0}x^x=1$
answered Dec 7 '18 at 11:24
José Carlos SantosJosé Carlos Santos
166k22132235
$endgroup$
$begingroup$
If I use it it seems like the limit is infinity. What is my mistake?
$endgroup$
– user519955
Dec 7 '18 at 12:02
$begingroup$
$lim_{(x,y)to(1,1)} frac{(x-y)^{(x-y)}} {(x-y)}$ = $lim_{(x,y)to(1,1)} (x-y)^{(x-y)}. lim_{(x,y)to(1,1)} frac{1} {(x-y)}$ $to$ $1. infty$
$endgroup$
– user519955
Dec 7 '18 at 12:04
2
$begingroup$
I didn't write them.
$endgroup$
– José Carlos Santos
Dec 7 '18 at 12:11
1
$begingroup$
@JoséCarlosSantos Divergent limits are not said to exist. A limit exists if it is real, finite and unique. Unless otherwise stated, existence of a limit is checked in the real numbers, not the extended real line.
$endgroup$
– Shubham Johri
Dec 7 '18 at 12:22
1
$begingroup$
So, for you the limit $lim_{xto+infty}x$ doesn't exist. I find it much more natural to say that it is $+infty$.
$endgroup$
– José Carlos Santos
Dec 7 '18 at 12:26
|
show 4 more comments
$begingroup$
Hint: $displaystylelim_{xto0}x^x=1$
answered Dec 7 '18 at 11:24
José Carlos SantosJosé Carlos Santos
166k22132235
$endgroup$
$begingroup$
If I use it it seems like the limit is infinity. What is my mistake?
$endgroup$
– user519955
Dec 7 '18 at 12:02
$begingroup$
$lim_{(x,y)to(1,1)} frac{(x-y)^{(x-y)}} {(x-y)}$ = $lim_{(x,y)to(1,1)} (x-y)^{(x-y)}. lim_{(x,y)to(1,1)} frac{1} {(x-y)}$ $to$ $1. infty$
$endgroup$
– user519955
Dec 7 '18 at 12:04
2
$begingroup$
I didn't write them.
$endgroup$
– José Carlos Santos
Dec 7 '18 at 12:11
1
$begingroup$
@JoséCarlosSantos Divergent limits are not said to exist. A limit exists if it is real, finite and unique. Unless otherwise stated, existence of a limit is checked in the real numbers, not the extended real line.
$endgroup$
– Shubham Johri
Dec 7 '18 at 12:22
1
$begingroup$
So, for you the limit $lim_{xto+infty}x$ doesn't exist. I find it much more natural to say that it is $+infty$.
$endgroup$
– José Carlos Santos
Dec 7 '18 at 12:26
|
show 4 more comments
$begingroup$
Hint: $displaystylelim_{xto0}x^x=1$
answered Dec 7 '18 at 11:24
José Carlos SantosJosé Carlos Santos
166k22132235
$endgroup$
Hint: $displaystylelim_{xto0}x^x=1$
answered Dec 7 '18 at 11:24
José Carlos SantosJosé Carlos Santos
166k22132235
answered Dec 7 '18 at 11:24
José Carlos SantosJosé Carlos Santos
166k22132235
answered Dec 7 '18 at 11:24
José Carlos SantosJosé Carlos Santos
166k22132235
answered Dec 7 '18 at 11:24
José Carlos SantosJosé Carlos Santos
166k22132235
166k22132235
$begingroup$
If I use it it seems like the limit is infinity. What is my mistake?
$endgroup$
– user519955
Dec 7 '18 at 12:02
$begingroup$
$lim_{(x,y)to(1,1)} frac{(x-y)^{(x-y)}} {(x-y)}$ = $lim_{(x,y)to(1,1)} (x-y)^{(x-y)}. lim_{(x,y)to(1,1)} frac{1} {(x-y)}$ $to$ $1. infty$
$endgroup$
– user519955
Dec 7 '18 at 12:04
2
$begingroup$
I didn't write them.
$endgroup$
– José Carlos Santos
Dec 7 '18 at 12:11
1
$begingroup$
@JoséCarlosSantos Divergent limits are not said to exist. A limit exists if it is real, finite and unique. Unless otherwise stated, existence of a limit is checked in the real numbers, not the extended real line.
$endgroup$
– Shubham Johri
Dec 7 '18 at 12:22
1
$begingroup$
So, for you the limit $lim_{xto+infty}x$ doesn't exist. I find it much more natural to say that it is $+infty$.
$endgroup$
– José Carlos Santos
Dec 7 '18 at 12:26
|
show 4 more comments
$begingroup$
If I use it it seems like the limit is infinity. What is my mistake?
$endgroup$
– user519955
Dec 7 '18 at 12:02
$begingroup$
$lim_{(x,y)to(1,1)} frac{(x-y)^{(x-y)}} {(x-y)}$ = $lim_{(x,y)to(1,1)} (x-y)^{(x-y)}. lim_{(x,y)to(1,1)} frac{1} {(x-y)}$ $to$ $1. infty$
$endgroup$
– user519955
Dec 7 '18 at 12:04
2
$begingroup$
I didn't write them.
$endgroup$
– José Carlos Santos
Dec 7 '18 at 12:11
1
$begingroup$
@JoséCarlosSantos Divergent limits are not said to exist. A limit exists if it is real, finite and unique. Unless otherwise stated, existence of a limit is checked in the real numbers, not the extended real line.
$endgroup$
– Shubham Johri
Dec 7 '18 at 12:22
1
$begingroup$
So, for you the limit $lim_{xto+infty}x$ doesn't exist. I find it much more natural to say that it is $+infty$.
$endgroup$
– José Carlos Santos
Dec 7 '18 at 12:26
$begingroup$
If I use it it seems like the limit is infinity. What is my mistake?
$endgroup$
– user519955
Dec 7 '18 at 12:02
$begingroup$
If I use it it seems like the limit is infinity. What is my mistake?
$endgroup$
– user519955
Dec 7 '18 at 12:02
$begingroup$
$lim_{(x,y)to(1,1)} frac{(x-y)^{(x-y)}} {(x-y)}$ = $lim_{(x,y)to(1,1)} (x-y)^{(x-y)}. lim_{(x,y)to(1,1)} frac{1} {(x-y)}$ $to$ $1. infty$
$endgroup$
– user519955
Dec 7 '18 at 12:04
$begingroup$
$lim_{(x,y)to(1,1)} frac{(x-y)^{(x-y)}} {(x-y)}$ = $lim_{(x,y)to(1,1)} (x-y)^{(x-y)}. lim_{(x,y)to(1,1)} frac{1} {(x-y)}$ $to$ $1. infty$
$endgroup$
– user519955
Dec 7 '18 at 12:04
2
2
$begingroup$
I didn't write them.
$endgroup$
– José Carlos Santos
Dec 7 '18 at 12:11
$begingroup$
I didn't write them.
$endgroup$
– José Carlos Santos
Dec 7 '18 at 12:11
1
1
$begingroup$
@JoséCarlosSantos Divergent limits are not said to exist. A limit exists if it is real, finite and unique. Unless otherwise stated, existence of a limit is checked in the real numbers, not the extended real line.
$endgroup$
– Shubham Johri
Dec 7 '18 at 12:22
$begingroup$
@JoséCarlosSantos Divergent limits are not said to exist. A limit exists if it is real, finite and unique. Unless otherwise stated, existence of a limit is checked in the real numbers, not the extended real line.
$endgroup$
– Shubham Johri
Dec 7 '18 at 12:22
1
1
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So, for you the limit $lim_{xto+infty}x$ doesn't exist. I find it much more natural to say that it is $+infty$.
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– José Carlos Santos
Dec 7 '18 at 12:26
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So, for you the limit $lim_{xto+infty}x$ doesn't exist. I find it much more natural to say that it is $+infty$.
$endgroup$
– José Carlos Santos
Dec 7 '18 at 12:26
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show 4 more comments
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No, this limit doesn't exist. Chose the path $y=x$. Alternatively, let $x-y=m$. Then the question is just about evaluating $lim_{mto0} frac{m^m}m=lim_{mto0}m^{m-1}$ which doesn't exist.
answered Dec 7 '18 at 11:29
Shubham JohriShubham Johri
5,204718
$endgroup$
$begingroup$
Come on, you can't choose the path $y=x$...
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– dallonsi
Dec 7 '18 at 11:51
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Why can't we chose $y=x$? It is pretty much a part of the domain.
$endgroup$
– Shubham Johri
Dec 7 '18 at 11:53
$begingroup$
if $y=x$ then you make a division by $0$ ! No matter what your numerator is, you still write a division by zero.
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– dallonsi
Dec 7 '18 at 11:54
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That is actually the point.
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– Shubham Johri
Dec 7 '18 at 11:55
1
$begingroup$
@gimusi $infty$ is not an object, it's a concept. A limit exists if it is equal to some number (not $infty$). Do your research online.
$endgroup$
– Algebear
Dec 7 '18 at 12:24
|
show 15 more comments
$begingroup$
No, this limit doesn't exist. Chose the path $y=x$. Alternatively, let $x-y=m$. Then the question is just about evaluating $lim_{mto0} frac{m^m}m=lim_{mto0}m^{m-1}$ which doesn't exist.
answered Dec 7 '18 at 11:29
Shubham JohriShubham Johri
5,204718
$endgroup$
$begingroup$
Come on, you can't choose the path $y=x$...
$endgroup$
– dallonsi
Dec 7 '18 at 11:51
$begingroup$
Why can't we chose $y=x$? It is pretty much a part of the domain.
$endgroup$
– Shubham Johri
Dec 7 '18 at 11:53
$begingroup$
if $y=x$ then you make a division by $0$ ! No matter what your numerator is, you still write a division by zero.
$endgroup$
– dallonsi
Dec 7 '18 at 11:54
$begingroup$
That is actually the point.
$endgroup$
– Shubham Johri
Dec 7 '18 at 11:55
1
$begingroup$
@gimusi $infty$ is not an object, it's a concept. A limit exists if it is equal to some number (not $infty$). Do your research online.
$endgroup$
– Algebear
Dec 7 '18 at 12:24
|
show 15 more comments
$begingroup$
No, this limit doesn't exist. Chose the path $y=x$. Alternatively, let $x-y=m$. Then the question is just about evaluating $lim_{mto0} frac{m^m}m=lim_{mto0}m^{m-1}$ which doesn't exist.
answered Dec 7 '18 at 11:29
Shubham JohriShubham Johri
5,204718
$endgroup$
No, this limit doesn't exist. Chose the path $y=x$. Alternatively, let $x-y=m$. Then the question is just about evaluating $lim_{mto0} frac{m^m}m=lim_{mto0}m^{m-1}$ which doesn't exist.
answered Dec 7 '18 at 11:29
Shubham JohriShubham Johri
5,204718
answered Dec 7 '18 at 11:29
Shubham JohriShubham Johri
5,204718
answered Dec 7 '18 at 11:29
Shubham JohriShubham Johri
5,204718
answered Dec 7 '18 at 11:29
Shubham JohriShubham Johri
5,204718
5,204718
$begingroup$
Come on, you can't choose the path $y=x$...
$endgroup$
– dallonsi
Dec 7 '18 at 11:51
$begingroup$
Why can't we chose $y=x$? It is pretty much a part of the domain.
$endgroup$
– Shubham Johri
Dec 7 '18 at 11:53
$begingroup$
if $y=x$ then you make a division by $0$ ! No matter what your numerator is, you still write a division by zero.
$endgroup$
– dallonsi
Dec 7 '18 at 11:54
$begingroup$
That is actually the point.
$endgroup$
– Shubham Johri
Dec 7 '18 at 11:55
1
$begingroup$
@gimusi $infty$ is not an object, it's a concept. A limit exists if it is equal to some number (not $infty$). Do your research online.
$endgroup$
– Algebear
Dec 7 '18 at 12:24
|
show 15 more comments
$begingroup$
Come on, you can't choose the path $y=x$...
$endgroup$
– dallonsi
Dec 7 '18 at 11:51
$begingroup$
Why can't we chose $y=x$? It is pretty much a part of the domain.
$endgroup$
– Shubham Johri
Dec 7 '18 at 11:53
$begingroup$
if $y=x$ then you make a division by $0$ ! No matter what your numerator is, you still write a division by zero.
$endgroup$
– dallonsi
Dec 7 '18 at 11:54
$begingroup$
That is actually the point.
$endgroup$
– Shubham Johri
Dec 7 '18 at 11:55
1
$begingroup$
@gimusi $infty$ is not an object, it's a concept. A limit exists if it is equal to some number (not $infty$). Do your research online.
$endgroup$
– Algebear
Dec 7 '18 at 12:24
$begingroup$
Come on, you can't choose the path $y=x$...
$endgroup$
– dallonsi
Dec 7 '18 at 11:51
$begingroup$
Come on, you can't choose the path $y=x$...
$endgroup$
– dallonsi
Dec 7 '18 at 11:51
$begingroup$
Why can't we chose $y=x$? It is pretty much a part of the domain.
$endgroup$
– Shubham Johri
Dec 7 '18 at 11:53
$begingroup$
Why can't we chose $y=x$? It is pretty much a part of the domain.
$endgroup$
– Shubham Johri
Dec 7 '18 at 11:53
$begingroup$
if $y=x$ then you make a division by $0$ ! No matter what your numerator is, you still write a division by zero.
$endgroup$
– dallonsi
Dec 7 '18 at 11:54
$begingroup$
if $y=x$ then you make a division by $0$ ! No matter what your numerator is, you still write a division by zero.
$endgroup$
– dallonsi
Dec 7 '18 at 11:54
$begingroup$
That is actually the point.
$endgroup$
– Shubham Johri
Dec 7 '18 at 11:55
$begingroup$
That is actually the point.
$endgroup$
– Shubham Johri
Dec 7 '18 at 11:55
1
1
$begingroup$
@gimusi $infty$ is not an object, it's a concept. A limit exists if it is equal to some number (not $infty$). Do your research online.
$endgroup$
– Algebear
Dec 7 '18 at 12:24
$begingroup$
@gimusi $infty$ is not an object, it's a concept. A limit exists if it is equal to some number (not $infty$). Do your research online.
$endgroup$
– Algebear
Dec 7 '18 at 12:24
|
show 15 more comments
$begingroup$
This limit does not exist.
Let $varepsilon in mathbb{R}^*_+$.
Take $x=1 + varepsilon$ and $y= 1$.
Then rewrite the quotient as :
$$varepsilon^{varepsilon -1} = e^{(varepsilon -1) ln (varepsilon)}.$$
Since $lim_{varepsilon to 0^+} varepsilon ln (varepsilon) =0$ and $lim_{varepsilon to 0^+} - ln (varepsilon) = +infty$, then the limit of $frac{(x-y)^{(x-y)}} {(x-y)}$ is $+infty$.
Now you want to apply the same trick with $-varepsilon$ to get a different limit. The thing is that the power of a negative number might be not be a real number (take for example the square root of $-2$). You can get rid off this problem by using complexe numbers.
Thus, take now $x = 1 - varepsilon$ and $y=1$. The quotient is now:
$$ frac{(-varepsilon)^{-varepsilon}}{-varepsilon} = e^{-ifrac{pi}{2}varepsilon}frac{varepsilon^{-varepsilon}}{-varepsilon} = e^{-ifrac{pi}{2}varepsilon}frac{1}{-varepsilon^{varepsilon + 1}} $$
from here you see that $e^{-ifrac{pi}{2}varepsilon}$ tends to 1 and the denominator $-varepsilon^{varepsilon + 1} $ goes to $0^{-}$. Thus the limit is now $-infty$.
Since you can find two "paths" $(x,y)$ which goes to $1$ and which give different limits, the limit $lim_{(x,y)to(1,1)} frac{(x-y)^{(x-y)}} {(x-y)}$ does not exist.
answered Dec 7 '18 at 11:49
dallonsidallonsi
1187
$endgroup$
$begingroup$
I think that your conclusion is wrong, the limit exists and it is $+infty$.
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– gimusi
Dec 7 '18 at 12:13
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Huw do you define $(-varepsilon)^{-varepsilon}$?
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– gimusi
Dec 7 '18 at 12:38
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well if $x,y in mathbb{R}$, you can always say that $x^y =|x|^y times e^{iarg(x) times y}$.
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– dallonsi
Dec 7 '18 at 13:17
$begingroup$
another way to say it is : $lim_{(x,y) to (1,1)} (x-y)^{(x-y)} = 1$, right ? So now, if you divide it by $(x-y)$, you can find a path that leads the quotient to $+infty$ (if $(x-y) to 0^+$) or $-infty$ (if $(x-y) to 0^{-}$). @gimusi, do you agree ?
$endgroup$
– dallonsi
Dec 7 '18 at 13:20
add a comment |
$begingroup$
This limit does not exist.
Let $varepsilon in mathbb{R}^*_+$.
Take $x=1 + varepsilon$ and $y= 1$.
Then rewrite the quotient as :
$$varepsilon^{varepsilon -1} = e^{(varepsilon -1) ln (varepsilon)}.$$
Since $lim_{varepsilon to 0^+} varepsilon ln (varepsilon) =0$ and $lim_{varepsilon to 0^+} - ln (varepsilon) = +infty$, then the limit of $frac{(x-y)^{(x-y)}} {(x-y)}$ is $+infty$.
Now you want to apply the same trick with $-varepsilon$ to get a different limit. The thing is that the power of a negative number might be not be a real number (take for example the square root of $-2$). You can get rid off this problem by using complexe numbers.
Thus, take now $x = 1 - varepsilon$ and $y=1$. The quotient is now:
$$ frac{(-varepsilon)^{-varepsilon}}{-varepsilon} = e^{-ifrac{pi}{2}varepsilon}frac{varepsilon^{-varepsilon}}{-varepsilon} = e^{-ifrac{pi}{2}varepsilon}frac{1}{-varepsilon^{varepsilon + 1}} $$
from here you see that $e^{-ifrac{pi}{2}varepsilon}$ tends to 1 and the denominator $-varepsilon^{varepsilon + 1} $ goes to $0^{-}$. Thus the limit is now $-infty$.
Since you can find two "paths" $(x,y)$ which goes to $1$ and which give different limits, the limit $lim_{(x,y)to(1,1)} frac{(x-y)^{(x-y)}} {(x-y)}$ does not exist.
answered Dec 7 '18 at 11:49
dallonsidallonsi
1187
$endgroup$
$begingroup$
I think that your conclusion is wrong, the limit exists and it is $+infty$.
$endgroup$
– gimusi
Dec 7 '18 at 12:13
$begingroup$
Huw do you define $(-varepsilon)^{-varepsilon}$?
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– gimusi
Dec 7 '18 at 12:38
$begingroup$
well if $x,y in mathbb{R}$, you can always say that $x^y =|x|^y times e^{iarg(x) times y}$.
$endgroup$
– dallonsi
Dec 7 '18 at 13:17
$begingroup$
another way to say it is : $lim_{(x,y) to (1,1)} (x-y)^{(x-y)} = 1$, right ? So now, if you divide it by $(x-y)$, you can find a path that leads the quotient to $+infty$ (if $(x-y) to 0^+$) or $-infty$ (if $(x-y) to 0^{-}$). @gimusi, do you agree ?
$endgroup$
– dallonsi
Dec 7 '18 at 13:20
add a comment |
$begingroup$
This limit does not exist.
Let $varepsilon in mathbb{R}^*_+$.
Take $x=1 + varepsilon$ and $y= 1$.
Then rewrite the quotient as :
$$varepsilon^{varepsilon -1} = e^{(varepsilon -1) ln (varepsilon)}.$$
Since $lim_{varepsilon to 0^+} varepsilon ln (varepsilon) =0$ and $lim_{varepsilon to 0^+} - ln (varepsilon) = +infty$, then the limit of $frac{(x-y)^{(x-y)}} {(x-y)}$ is $+infty$.
Now you want to apply the same trick with $-varepsilon$ to get a different limit. The thing is that the power of a negative number might be not be a real number (take for example the square root of $-2$). You can get rid off this problem by using complexe numbers.
Thus, take now $x = 1 - varepsilon$ and $y=1$. The quotient is now:
$$ frac{(-varepsilon)^{-varepsilon}}{-varepsilon} = e^{-ifrac{pi}{2}varepsilon}frac{varepsilon^{-varepsilon}}{-varepsilon} = e^{-ifrac{pi}{2}varepsilon}frac{1}{-varepsilon^{varepsilon + 1}} $$
from here you see that $e^{-ifrac{pi}{2}varepsilon}$ tends to 1 and the denominator $-varepsilon^{varepsilon + 1} $ goes to $0^{-}$. Thus the limit is now $-infty$.
Since you can find two "paths" $(x,y)$ which goes to $1$ and which give different limits, the limit $lim_{(x,y)to(1,1)} frac{(x-y)^{(x-y)}} {(x-y)}$ does not exist.
answered Dec 7 '18 at 11:49
dallonsidallonsi
1187
$endgroup$
This limit does not exist.
Let $varepsilon in mathbb{R}^*_+$.
Take $x=1 + varepsilon$ and $y= 1$.
Then rewrite the quotient as :
$$varepsilon^{varepsilon -1} = e^{(varepsilon -1) ln (varepsilon)}.$$
Since $lim_{varepsilon to 0^+} varepsilon ln (varepsilon) =0$ and $lim_{varepsilon to 0^+} - ln (varepsilon) = +infty$, then the limit of $frac{(x-y)^{(x-y)}} {(x-y)}$ is $+infty$.
Now you want to apply the same trick with $-varepsilon$ to get a different limit. The thing is that the power of a negative number might be not be a real number (take for example the square root of $-2$). You can get rid off this problem by using complexe numbers.
Thus, take now $x = 1 - varepsilon$ and $y=1$. The quotient is now:
$$ frac{(-varepsilon)^{-varepsilon}}{-varepsilon} = e^{-ifrac{pi}{2}varepsilon}frac{varepsilon^{-varepsilon}}{-varepsilon} = e^{-ifrac{pi}{2}varepsilon}frac{1}{-varepsilon^{varepsilon + 1}} $$
from here you see that $e^{-ifrac{pi}{2}varepsilon}$ tends to 1 and the denominator $-varepsilon^{varepsilon + 1} $ goes to $0^{-}$. Thus the limit is now $-infty$.
Since you can find two "paths" $(x,y)$ which goes to $1$ and which give different limits, the limit $lim_{(x,y)to(1,1)} frac{(x-y)^{(x-y)}} {(x-y)}$ does not exist.
answered Dec 7 '18 at 11:49
dallonsidallonsi
1187
answered Dec 7 '18 at 11:49
dallonsidallonsi
1187
answered Dec 7 '18 at 11:49
dallonsidallonsi
1187
answered Dec 7 '18 at 11:49
dallonsidallonsi
1187
1187
$begingroup$
I think that your conclusion is wrong, the limit exists and it is $+infty$.
$endgroup$
– gimusi
Dec 7 '18 at 12:13
$begingroup$
Huw do you define $(-varepsilon)^{-varepsilon}$?
$endgroup$
– gimusi
Dec 7 '18 at 12:38
$begingroup$
well if $x,y in mathbb{R}$, you can always say that $x^y =|x|^y times e^{iarg(x) times y}$.
$endgroup$
– dallonsi
Dec 7 '18 at 13:17
$begingroup$
another way to say it is : $lim_{(x,y) to (1,1)} (x-y)^{(x-y)} = 1$, right ? So now, if you divide it by $(x-y)$, you can find a path that leads the quotient to $+infty$ (if $(x-y) to 0^+$) or $-infty$ (if $(x-y) to 0^{-}$). @gimusi, do you agree ?
$endgroup$
– dallonsi
Dec 7 '18 at 13:20
add a comment |
$begingroup$
I think that your conclusion is wrong, the limit exists and it is $+infty$.
$endgroup$
– gimusi
Dec 7 '18 at 12:13
$begingroup$
Huw do you define $(-varepsilon)^{-varepsilon}$?
$endgroup$
– gimusi
Dec 7 '18 at 12:38
$begingroup$
well if $x,y in mathbb{R}$, you can always say that $x^y =|x|^y times e^{iarg(x) times y}$.
$endgroup$
– dallonsi
Dec 7 '18 at 13:17
$begingroup$
another way to say it is : $lim_{(x,y) to (1,1)} (x-y)^{(x-y)} = 1$, right ? So now, if you divide it by $(x-y)$, you can find a path that leads the quotient to $+infty$ (if $(x-y) to 0^+$) or $-infty$ (if $(x-y) to 0^{-}$). @gimusi, do you agree ?
$endgroup$
– dallonsi
Dec 7 '18 at 13:20
$begingroup$
I think that your conclusion is wrong, the limit exists and it is $+infty$.
$endgroup$
– gimusi
Dec 7 '18 at 12:13
$begingroup$
I think that your conclusion is wrong, the limit exists and it is $+infty$.
$endgroup$
– gimusi
Dec 7 '18 at 12:13
$begingroup$
Huw do you define $(-varepsilon)^{-varepsilon}$?
$endgroup$
– gimusi
Dec 7 '18 at 12:38
$begingroup$
Huw do you define $(-varepsilon)^{-varepsilon}$?
$endgroup$
– gimusi
Dec 7 '18 at 12:38
$begingroup$
well if $x,y in mathbb{R}$, you can always say that $x^y =|x|^y times e^{iarg(x) times y}$.
$endgroup$
– dallonsi
Dec 7 '18 at 13:17
$begingroup$
well if $x,y in mathbb{R}$, you can always say that $x^y =|x|^y times e^{iarg(x) times y}$.
$endgroup$
– dallonsi
Dec 7 '18 at 13:17
$begingroup$
another way to say it is : $lim_{(x,y) to (1,1)} (x-y)^{(x-y)} = 1$, right ? So now, if you divide it by $(x-y)$, you can find a path that leads the quotient to $+infty$ (if $(x-y) to 0^+$) or $-infty$ (if $(x-y) to 0^{-}$). @gimusi, do you agree ?
$endgroup$
– dallonsi
Dec 7 '18 at 13:20
$begingroup$
another way to say it is : $lim_{(x,y) to (1,1)} (x-y)^{(x-y)} = 1$, right ? So now, if you divide it by $(x-y)$, you can find a path that leads the quotient to $+infty$ (if $(x-y) to 0^+$) or $-infty$ (if $(x-y) to 0^{-}$). @gimusi, do you agree ?
$endgroup$
– dallonsi
Dec 7 '18 at 13:20
add a comment |
$begingroup$
We have that by $x-y =t to 0^+$ the given limit reduces to
$$lim_{(x,y)to(1,1)} frac{(x-y)^{(x-y)}} {(x-y)}=lim_{tto 0^+} frac{t^t} {t}to infty$$
indeed the points $(x-y)<0$ are not included within the domain of definition for the function that is $$D={(x,y)inmathbb{R^2}:x-y>0}$$
Refer also to the related
- domain of $x^x$
Edit for a remark
As noticed by Jyrky Lahtonen in the comment "some sources would insist that for the limit to exist the function should be defined in a punctured neighborhood of $(1,1)$. Other sources would never worry about points outside the domain of definition, in which case you are restricted to the half-plane $x>y$".
That important issue has been deeply discussed here
- What is $lim_{x to 0}frac{sin(frac 1x)}{sin (frac 1 x)}$ ? Does it exist?
My observation is that not all sources have to be considered at the same level. Using the first definition is a reasonable way to deal with limits at a lower (high school) level whereas, in a more advanced context, the second more general definition (according to Rudin) should be adopted.
Moreover, note that, according to the first definition, the discussion of limits would reduce, as in the present case, to a (more or less) trivial determination on the domain of definition for functions which is of course a very different topic.
Therefore, when we deal with limits, I strongly suggest to refer to the more general definition excluding points outside the domain of definition for the function considered.
answered Dec 7 '18 at 12:12
gimusigimusi
93k84494
$endgroup$
1
$begingroup$
Why aren't $(x-y) lt 0$ included in the domain? Domain is all $mathbb R^2$
$endgroup$
– user519955
Dec 7 '18 at 12:17
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@user519955 The key point is exactly that one. For which values is $f(x)=x^x$ defined?
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– gimusi
Dec 7 '18 at 12:18
$begingroup$
@user519955 I suggest to refer to that OP and notably to the answer HERE. It often a point of counfusion. Do not hesitate to ask for any clarifiaction on that.
$endgroup$
– gimusi
Dec 7 '18 at 12:24
1
$begingroup$
I will glance it. But domain seems like $mathbb R - {0} cup {1/(2n | nin mathbb N}$ for $x^x$ not $x>0$ to me. For example $(-2)^{(-2)}=1/4$
$endgroup$
– user519955
Dec 7 '18 at 12:48
1
$begingroup$
@user519955 That’s a good point, I agree that the expression $x^x$ is defined for some values as also for example $(-1/3)^{-1/3}$ but in the calculus context the domain for the function $f(x)=x^x$ is defined only for $x>0$. Refer to that OP for a discussion on that.
$endgroup$
– gimusi
Dec 7 '18 at 13:00
add a comment |
$begingroup$
We have that by $x-y =t to 0^+$ the given limit reduces to
$$lim_{(x,y)to(1,1)} frac{(x-y)^{(x-y)}} {(x-y)}=lim_{tto 0^+} frac{t^t} {t}to infty$$
indeed the points $(x-y)<0$ are not included within the domain of definition for the function that is $$D={(x,y)inmathbb{R^2}:x-y>0}$$
Refer also to the related
- domain of $x^x$
Edit for a remark
As noticed by Jyrky Lahtonen in the comment "some sources would insist that for the limit to exist the function should be defined in a punctured neighborhood of $(1,1)$. Other sources would never worry about points outside the domain of definition, in which case you are restricted to the half-plane $x>y$".
That important issue has been deeply discussed here
- What is $lim_{x to 0}frac{sin(frac 1x)}{sin (frac 1 x)}$ ? Does it exist?
My observation is that not all sources have to be considered at the same level. Using the first definition is a reasonable way to deal with limits at a lower (high school) level whereas, in a more advanced context, the second more general definition (according to Rudin) should be adopted.
Moreover, note that, according to the first definition, the discussion of limits would reduce, as in the present case, to a (more or less) trivial determination on the domain of definition for functions which is of course a very different topic.
Therefore, when we deal with limits, I strongly suggest to refer to the more general definition excluding points outside the domain of definition for the function considered.
answered Dec 7 '18 at 12:12
gimusigimusi
93k84494
$endgroup$
1
$begingroup$
Why aren't $(x-y) lt 0$ included in the domain? Domain is all $mathbb R^2$
$endgroup$
– user519955
Dec 7 '18 at 12:17
$begingroup$
@user519955 The key point is exactly that one. For which values is $f(x)=x^x$ defined?
$endgroup$
– gimusi
Dec 7 '18 at 12:18
$begingroup$
@user519955 I suggest to refer to that OP and notably to the answer HERE. It often a point of counfusion. Do not hesitate to ask for any clarifiaction on that.
$endgroup$
– gimusi
Dec 7 '18 at 12:24
1
$begingroup$
I will glance it. But domain seems like $mathbb R - {0} cup {1/(2n | nin mathbb N}$ for $x^x$ not $x>0$ to me. For example $(-2)^{(-2)}=1/4$
$endgroup$
– user519955
Dec 7 '18 at 12:48
1
$begingroup$
@user519955 That’s a good point, I agree that the expression $x^x$ is defined for some values as also for example $(-1/3)^{-1/3}$ but in the calculus context the domain for the function $f(x)=x^x$ is defined only for $x>0$. Refer to that OP for a discussion on that.
$endgroup$
– gimusi
Dec 7 '18 at 13:00
add a comment |
$begingroup$
We have that by $x-y =t to 0^+$ the given limit reduces to
$$lim_{(x,y)to(1,1)} frac{(x-y)^{(x-y)}} {(x-y)}=lim_{tto 0^+} frac{t^t} {t}to infty$$
indeed the points $(x-y)<0$ are not included within the domain of definition for the function that is $$D={(x,y)inmathbb{R^2}:x-y>0}$$
Refer also to the related
- domain of $x^x$
Edit for a remark
As noticed by Jyrky Lahtonen in the comment "some sources would insist that for the limit to exist the function should be defined in a punctured neighborhood of $(1,1)$. Other sources would never worry about points outside the domain of definition, in which case you are restricted to the half-plane $x>y$".
That important issue has been deeply discussed here
- What is $lim_{x to 0}frac{sin(frac 1x)}{sin (frac 1 x)}$ ? Does it exist?
My observation is that not all sources have to be considered at the same level. Using the first definition is a reasonable way to deal with limits at a lower (high school) level whereas, in a more advanced context, the second more general definition (according to Rudin) should be adopted.
Moreover, note that, according to the first definition, the discussion of limits would reduce, as in the present case, to a (more or less) trivial determination on the domain of definition for functions which is of course a very different topic.
Therefore, when we deal with limits, I strongly suggest to refer to the more general definition excluding points outside the domain of definition for the function considered.
answered Dec 7 '18 at 12:12
gimusigimusi
93k84494
$endgroup$
We have that by $x-y =t to 0^+$ the given limit reduces to
$$lim_{(x,y)to(1,1)} frac{(x-y)^{(x-y)}} {(x-y)}=lim_{tto 0^+} frac{t^t} {t}to infty$$
indeed the points $(x-y)<0$ are not included within the domain of definition for the function that is $$D={(x,y)inmathbb{R^2}:x-y>0}$$
Refer also to the related
- domain of $x^x$
Edit for a remark
As noticed by Jyrky Lahtonen in the comment "some sources would insist that for the limit to exist the function should be defined in a punctured neighborhood of $(1,1)$. Other sources would never worry about points outside the domain of definition, in which case you are restricted to the half-plane $x>y$".
That important issue has been deeply discussed here
- What is $lim_{x to 0}frac{sin(frac 1x)}{sin (frac 1 x)}$ ? Does it exist?
My observation is that not all sources have to be considered at the same level. Using the first definition is a reasonable way to deal with limits at a lower (high school) level whereas, in a more advanced context, the second more general definition (according to Rudin) should be adopted.
Moreover, note that, according to the first definition, the discussion of limits would reduce, as in the present case, to a (more or less) trivial determination on the domain of definition for functions which is of course a very different topic.
Therefore, when we deal with limits, I strongly suggest to refer to the more general definition excluding points outside the domain of definition for the function considered.
answered Dec 7 '18 at 12:12
gimusigimusi
93k84494
edited Dec 7 '18 at 18:44
answered Dec 7 '18 at 12:12
gimusigimusi
93k84494
answered Dec 7 '18 at 12:12
gimusigimusi
93k84494
answered Dec 7 '18 at 12:12
gimusigimusi
93k84494
93k84494
1
$begingroup$
Why aren't $(x-y) lt 0$ included in the domain? Domain is all $mathbb R^2$
$endgroup$
– user519955
Dec 7 '18 at 12:17
$begingroup$
@user519955 The key point is exactly that one. For which values is $f(x)=x^x$ defined?
$endgroup$
– gimusi
Dec 7 '18 at 12:18
$begingroup$
@user519955 I suggest to refer to that OP and notably to the answer HERE. It often a point of counfusion. Do not hesitate to ask for any clarifiaction on that.
$endgroup$
– gimusi
Dec 7 '18 at 12:24
1
$begingroup$
I will glance it. But domain seems like $mathbb R - {0} cup {1/(2n | nin mathbb N}$ for $x^x$ not $x>0$ to me. For example $(-2)^{(-2)}=1/4$
$endgroup$
– user519955
Dec 7 '18 at 12:48
1
$begingroup$
@user519955 That’s a good point, I agree that the expression $x^x$ is defined for some values as also for example $(-1/3)^{-1/3}$ but in the calculus context the domain for the function $f(x)=x^x$ is defined only for $x>0$. Refer to that OP for a discussion on that.
$endgroup$
– gimusi
Dec 7 '18 at 13:00
add a comment |
1
$begingroup$
Why aren't $(x-y) lt 0$ included in the domain? Domain is all $mathbb R^2$
$endgroup$
– user519955
Dec 7 '18 at 12:17
$begingroup$
@user519955 The key point is exactly that one. For which values is $f(x)=x^x$ defined?
$endgroup$
– gimusi
Dec 7 '18 at 12:18
$begingroup$
@user519955 I suggest to refer to that OP and notably to the answer HERE. It often a point of counfusion. Do not hesitate to ask for any clarifiaction on that.
$endgroup$
– gimusi
Dec 7 '18 at 12:24
1
$begingroup$
I will glance it. But domain seems like $mathbb R - {0} cup {1/(2n | nin mathbb N}$ for $x^x$ not $x>0$ to me. For example $(-2)^{(-2)}=1/4$
$endgroup$
– user519955
Dec 7 '18 at 12:48
1
$begingroup$
@user519955 That’s a good point, I agree that the expression $x^x$ is defined for some values as also for example $(-1/3)^{-1/3}$ but in the calculus context the domain for the function $f(x)=x^x$ is defined only for $x>0$. Refer to that OP for a discussion on that.
$endgroup$
– gimusi
Dec 7 '18 at 13:00
1
1
$begingroup$
Why aren't $(x-y) lt 0$ included in the domain? Domain is all $mathbb R^2$
$endgroup$
– user519955
Dec 7 '18 at 12:17
$begingroup$
Why aren't $(x-y) lt 0$ included in the domain? Domain is all $mathbb R^2$
$endgroup$
– user519955
Dec 7 '18 at 12:17
$begingroup$
@user519955 The key point is exactly that one. For which values is $f(x)=x^x$ defined?
$endgroup$
– gimusi
Dec 7 '18 at 12:18
$begingroup$
@user519955 The key point is exactly that one. For which values is $f(x)=x^x$ defined?
$endgroup$
– gimusi
Dec 7 '18 at 12:18
$begingroup$
@user519955 I suggest to refer to that OP and notably to the answer HERE. It often a point of counfusion. Do not hesitate to ask for any clarifiaction on that.
$endgroup$
– gimusi
Dec 7 '18 at 12:24
$begingroup$
@user519955 I suggest to refer to that OP and notably to the answer HERE. It often a point of counfusion. Do not hesitate to ask for any clarifiaction on that.
$endgroup$
– gimusi
Dec 7 '18 at 12:24
1
1
$begingroup$
I will glance it. But domain seems like $mathbb R - {0} cup {1/(2n | nin mathbb N}$ for $x^x$ not $x>0$ to me. For example $(-2)^{(-2)}=1/4$
$endgroup$
– user519955
Dec 7 '18 at 12:48
$begingroup$
I will glance it. But domain seems like $mathbb R - {0} cup {1/(2n | nin mathbb N}$ for $x^x$ not $x>0$ to me. For example $(-2)^{(-2)}=1/4$
$endgroup$
– user519955
Dec 7 '18 at 12:48
1
1
$begingroup$
@user519955 That’s a good point, I agree that the expression $x^x$ is defined for some values as also for example $(-1/3)^{-1/3}$ but in the calculus context the domain for the function $f(x)=x^x$ is defined only for $x>0$. Refer to that OP for a discussion on that.
$endgroup$
– gimusi
Dec 7 '18 at 13:00
$begingroup$
@user519955 That’s a good point, I agree that the expression $x^x$ is defined for some values as also for example $(-1/3)^{-1/3}$ but in the calculus context the domain for the function $f(x)=x^x$ is defined only for $x>0$. Refer to that OP for a discussion on that.
$endgroup$
– gimusi
Dec 7 '18 at 13:00
add a comment |
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$begingroup$
You need to consider paths within the domain of definition for the function that is $$D={(x,y)inmathbb{R^2}:x-y>0}$$
$endgroup$
– gimusi
Dec 7 '18 at 12:32
2
$begingroup$
Some sources would insist that for the limit to exist the function should be defined in a punctured neighborhood of $(1,1)$. Other sources would never worry about points outside the domain of definition, in which case you are restricted to the half-plane $x>y$. Check what is the definition in use in your book, please!
$endgroup$
– Jyrki Lahtonen
Dec 7 '18 at 13:16
1
$begingroup$
@JyrkiLahtonen We have already discussed that point recently. Not all sources are to be considered at the same level. Using the first definition is reasonable to a lower (high school) level whereas in a more advanced context the second definition (according to Rudin) should be adopted. Notably according to the first definiiton the interesting question on the present limit would become a trivial question on the domain of the function.
$endgroup$
– gimusi
Dec 7 '18 at 14:32
$begingroup$
Indeed, @gimusi. I do remember that discussion. I still think that it is important for the askers to be aware of this (and to explain). What you say about the question becoming trivial otherwise is, of course, valid.
$endgroup$
– Jyrki Lahtonen
Dec 7 '18 at 17:28