Show that $limlimits_{x to infty}frac{ln x}{x}=0$ from definition.
up vote
1
down vote
favorite
Show that $displaystylelim_{x to infty}frac{ln x}{x}=0$.
I know this is a simple application of the L'Hopital's rule, but can we also show this from the $displaystyleepsilon-delta$ definition?
I am stuck because while it is easy to find a lower bound for the denominator $x$, the numerator does not have an upper bound - it merely increases less fast than the denominator. Is there a way to manipulate the expression to get a bound?
calculus real-analysis limits limits-without-lhopital
edited Nov 16 at 11:55
Martin Sleziak
44.4k7115268
asked Nov 15 at 18:33
Bunbury
627
add a comment |
up vote
1
down vote
favorite
Show that $displaystylelim_{x to infty}frac{ln x}{x}=0$.
I know this is a simple application of the L'Hopital's rule, but can we also show this from the $displaystyleepsilon-delta$ definition?
I am stuck because while it is easy to find a lower bound for the denominator $x$, the numerator does not have an upper bound - it merely increases less fast than the denominator. Is there a way to manipulate the expression to get a bound?
calculus real-analysis limits limits-without-lhopital
edited Nov 16 at 11:55
Martin Sleziak
44.4k7115268
asked Nov 15 at 18:33
Bunbury
627
$ln(x)leq x$ for $x>0$ and show it for the function $frac{1}{x}$.
– Yadati Kiran
Nov 15 at 18:38
@Rustyn there is one for infinite limits
– gt6989b
Nov 15 at 18:38
@gt6989b many thanks, I'll have a look
– Rustyn
Nov 15 at 18:39
3
If you let $x = e^y$, then it's equivalent to show that $$ lim_{y to infty} frac{y}{e^y} = 0 $$
– Omnomnomnom
Nov 15 at 18:42
1
Well its just an immediate but non-obvious consequence of the inequality $log xleq x-1$ or if you prefer the simpler inequality $log x<x$. The non-obvious part is related to the fact that we need to replace $x$ with $sqrt{x} $ in above inequality to get $log x<2sqrt{x}$.
– Paramanand Singh
Nov 15 at 18:53
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Show that $displaystylelim_{x to infty}frac{ln x}{x}=0$.
I know this is a simple application of the L'Hopital's rule, but can we also show this from the $displaystyleepsilon-delta$ definition?
I am stuck because while it is easy to find a lower bound for the denominator $x$, the numerator does not have an upper bound - it merely increases less fast than the denominator. Is there a way to manipulate the expression to get a bound?
calculus real-analysis limits limits-without-lhopital
edited Nov 16 at 11:55
Martin Sleziak
44.4k7115268
asked Nov 15 at 18:33
Bunbury
627
Show that $displaystylelim_{x to infty}frac{ln x}{x}=0$.
I know this is a simple application of the L'Hopital's rule, but can we also show this from the $displaystyleepsilon-delta$ definition?
I am stuck because while it is easy to find a lower bound for the denominator $x$, the numerator does not have an upper bound - it merely increases less fast than the denominator. Is there a way to manipulate the expression to get a bound?
calculus real-analysis limits limits-without-lhopital
calculus real-analysis limits limits-without-lhopital
edited Nov 16 at 11:55
Martin Sleziak
44.4k7115268
asked Nov 15 at 18:33
Bunbury
627
edited Nov 16 at 11:55
Martin Sleziak
44.4k7115268
asked Nov 15 at 18:33
Bunbury
627
edited Nov 16 at 11:55
Martin Sleziak
44.4k7115268
edited Nov 16 at 11:55
Martin Sleziak
44.4k7115268
edited Nov 16 at 11:55
Martin Sleziak
44.4k7115268
44.4k7115268
asked Nov 15 at 18:33
Bunbury
627
asked Nov 15 at 18:33
Bunbury
627
asked Nov 15 at 18:33
Bunbury
627
627
$ln(x)leq x$ for $x>0$ and show it for the function $frac{1}{x}$.
– Yadati Kiran
Nov 15 at 18:38
@Rustyn there is one for infinite limits
– gt6989b
Nov 15 at 18:38
@gt6989b many thanks, I'll have a look
– Rustyn
Nov 15 at 18:39
3
If you let $x = e^y$, then it's equivalent to show that $$ lim_{y to infty} frac{y}{e^y} = 0 $$
– Omnomnomnom
Nov 15 at 18:42
1
Well its just an immediate but non-obvious consequence of the inequality $log xleq x-1$ or if you prefer the simpler inequality $log x<x$. The non-obvious part is related to the fact that we need to replace $x$ with $sqrt{x} $ in above inequality to get $log x<2sqrt{x}$.
– Paramanand Singh
Nov 15 at 18:53
add a comment |
$ln(x)leq x$ for $x>0$ and show it for the function $frac{1}{x}$.
– Yadati Kiran
Nov 15 at 18:38
@Rustyn there is one for infinite limits
– gt6989b
Nov 15 at 18:38
@gt6989b many thanks, I'll have a look
– Rustyn
Nov 15 at 18:39
3
If you let $x = e^y$, then it's equivalent to show that $$ lim_{y to infty} frac{y}{e^y} = 0 $$
– Omnomnomnom
Nov 15 at 18:42
1
Well its just an immediate but non-obvious consequence of the inequality $log xleq x-1$ or if you prefer the simpler inequality $log x<x$. The non-obvious part is related to the fact that we need to replace $x$ with $sqrt{x} $ in above inequality to get $log x<2sqrt{x}$.
– Paramanand Singh
Nov 15 at 18:53
$ln(x)leq x$ for $x>0$ and show it for the function $frac{1}{x}$.
– Yadati Kiran
Nov 15 at 18:38
$ln(x)leq x$ for $x>0$ and show it for the function $frac{1}{x}$.
– Yadati Kiran
Nov 15 at 18:38
@Rustyn there is one for infinite limits
– gt6989b
Nov 15 at 18:38
@Rustyn there is one for infinite limits
– gt6989b
Nov 15 at 18:38
@gt6989b many thanks, I'll have a look
– Rustyn
Nov 15 at 18:39
@gt6989b many thanks, I'll have a look
– Rustyn
Nov 15 at 18:39
3
3
If you let $x = e^y$, then it's equivalent to show that $$ lim_{y to infty} frac{y}{e^y} = 0 $$
– Omnomnomnom
Nov 15 at 18:42
If you let $x = e^y$, then it's equivalent to show that $$ lim_{y to infty} frac{y}{e^y} = 0 $$
– Omnomnomnom
Nov 15 at 18:42
1
1
Well its just an immediate but non-obvious consequence of the inequality $log xleq x-1$ or if you prefer the simpler inequality $log x<x$. The non-obvious part is related to the fact that we need to replace $x$ with $sqrt{x} $ in above inequality to get $log x<2sqrt{x}$.
– Paramanand Singh
Nov 15 at 18:53
Well its just an immediate but non-obvious consequence of the inequality $log xleq x-1$ or if you prefer the simpler inequality $log x<x$. The non-obvious part is related to the fact that we need to replace $x$ with $sqrt{x} $ in above inequality to get $log x<2sqrt{x}$.
– Paramanand Singh
Nov 15 at 18:53
add a comment |
2 Answers
2
active
oldest
votes
up vote
6
down vote
accepted
Assuming you define the $ln(cdot) $ by the integral, then it is more or less enough to notice that
$$0leq int^x_1frac 1 t, dtleq int^x_1frac 1{ sqrt t} , dtleq int^x_0frac{1}{sqrt t} , dt$$
This leads to
$$left |frac{ln(x)} {x} right|leq left|frac{2sqrt{x}}{x} right|$$
The rest is straightforward.
answered Nov 15 at 18:39
Shashi
6,9191527
Ah yes! I forgot that's how we defined log function. Thank you!
– Bunbury
Nov 15 at 18:51
@Bunbury glad I could help, I made it a little bit easier now.
– Shashi
Nov 15 at 18:58
add a comment |
up vote
1
down vote
For all $x > 0, ln x < sqrt x$
$frac {ln x}{x} < frac {1}{sqrt x}$
When $N > frac {1}{epsilon^2},$ then $x>N implies frac {ln x}{x} < frac {1}{sqrt x} < epsilon$
answered Nov 15 at 18:53
Doug M
43k31752
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
6
down vote
accepted
Assuming you define the $ln(cdot) $ by the integral, then it is more or less enough to notice that
$$0leq int^x_1frac 1 t, dtleq int^x_1frac 1{ sqrt t} , dtleq int^x_0frac{1}{sqrt t} , dt$$
This leads to
$$left |frac{ln(x)} {x} right|leq left|frac{2sqrt{x}}{x} right|$$
The rest is straightforward.
answered Nov 15 at 18:39
Shashi
6,9191527
Ah yes! I forgot that's how we defined log function. Thank you!
– Bunbury
Nov 15 at 18:51
@Bunbury glad I could help, I made it a little bit easier now.
– Shashi
Nov 15 at 18:58
add a comment |
up vote
6
down vote
accepted
Assuming you define the $ln(cdot) $ by the integral, then it is more or less enough to notice that
$$0leq int^x_1frac 1 t, dtleq int^x_1frac 1{ sqrt t} , dtleq int^x_0frac{1}{sqrt t} , dt$$
This leads to
$$left |frac{ln(x)} {x} right|leq left|frac{2sqrt{x}}{x} right|$$
The rest is straightforward.
answered Nov 15 at 18:39
Shashi
6,9191527
Ah yes! I forgot that's how we defined log function. Thank you!
– Bunbury
Nov 15 at 18:51
@Bunbury glad I could help, I made it a little bit easier now.
– Shashi
Nov 15 at 18:58
add a comment |
up vote
6
down vote
accepted
up vote
6
down vote
accepted
Assuming you define the $ln(cdot) $ by the integral, then it is more or less enough to notice that
$$0leq int^x_1frac 1 t, dtleq int^x_1frac 1{ sqrt t} , dtleq int^x_0frac{1}{sqrt t} , dt$$
This leads to
$$left |frac{ln(x)} {x} right|leq left|frac{2sqrt{x}}{x} right|$$
The rest is straightforward.
answered Nov 15 at 18:39
Shashi
6,9191527
Assuming you define the $ln(cdot) $ by the integral, then it is more or less enough to notice that
$$0leq int^x_1frac 1 t, dtleq int^x_1frac 1{ sqrt t} , dtleq int^x_0frac{1}{sqrt t} , dt$$
This leads to
$$left |frac{ln(x)} {x} right|leq left|frac{2sqrt{x}}{x} right|$$
The rest is straightforward.
answered Nov 15 at 18:39
Shashi
6,9191527
edited Nov 15 at 18:56
answered Nov 15 at 18:39
Shashi
6,9191527
answered Nov 15 at 18:39
Shashi
6,9191527
answered Nov 15 at 18:39
Shashi
6,9191527
6,9191527
Ah yes! I forgot that's how we defined log function. Thank you!
– Bunbury
Nov 15 at 18:51
@Bunbury glad I could help, I made it a little bit easier now.
– Shashi
Nov 15 at 18:58
add a comment |
Ah yes! I forgot that's how we defined log function. Thank you!
– Bunbury
Nov 15 at 18:51
@Bunbury glad I could help, I made it a little bit easier now.
– Shashi
Nov 15 at 18:58
Ah yes! I forgot that's how we defined log function. Thank you!
– Bunbury
Nov 15 at 18:51
Ah yes! I forgot that's how we defined log function. Thank you!
– Bunbury
Nov 15 at 18:51
@Bunbury glad I could help, I made it a little bit easier now.
– Shashi
Nov 15 at 18:58
@Bunbury glad I could help, I made it a little bit easier now.
– Shashi
Nov 15 at 18:58
add a comment |
up vote
1
down vote
For all $x > 0, ln x < sqrt x$
$frac {ln x}{x} < frac {1}{sqrt x}$
When $N > frac {1}{epsilon^2},$ then $x>N implies frac {ln x}{x} < frac {1}{sqrt x} < epsilon$
answered Nov 15 at 18:53
Doug M
43k31752
add a comment |
up vote
1
down vote
For all $x > 0, ln x < sqrt x$
$frac {ln x}{x} < frac {1}{sqrt x}$
When $N > frac {1}{epsilon^2},$ then $x>N implies frac {ln x}{x} < frac {1}{sqrt x} < epsilon$
answered Nov 15 at 18:53
Doug M
43k31752
add a comment |
up vote
1
down vote
up vote
1
down vote
For all $x > 0, ln x < sqrt x$
$frac {ln x}{x} < frac {1}{sqrt x}$
When $N > frac {1}{epsilon^2},$ then $x>N implies frac {ln x}{x} < frac {1}{sqrt x} < epsilon$
answered Nov 15 at 18:53
Doug M
43k31752
For all $x > 0, ln x < sqrt x$
$frac {ln x}{x} < frac {1}{sqrt x}$
When $N > frac {1}{epsilon^2},$ then $x>N implies frac {ln x}{x} < frac {1}{sqrt x} < epsilon$
answered Nov 15 at 18:53
Doug M
43k31752
answered Nov 15 at 18:53
Doug M
43k31752
answered Nov 15 at 18:53
Doug M
43k31752
answered Nov 15 at 18:53
Doug M
43k31752
43k31752
add a comment |
add a comment |
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$ln(x)leq x$ for $x>0$ and show it for the function $frac{1}{x}$.
– Yadati Kiran
Nov 15 at 18:38
@Rustyn there is one for infinite limits
– gt6989b
Nov 15 at 18:38
@gt6989b many thanks, I'll have a look
– Rustyn
Nov 15 at 18:39
3
If you let $x = e^y$, then it's equivalent to show that $$ lim_{y to infty} frac{y}{e^y} = 0 $$
– Omnomnomnom
Nov 15 at 18:42
1
Well its just an immediate but non-obvious consequence of the inequality $log xleq x-1$ or if you prefer the simpler inequality $log x<x$. The non-obvious part is related to the fact that we need to replace $x$ with $sqrt{x} $ in above inequality to get $log x<2sqrt{x}$.
– Paramanand Singh
Nov 15 at 18:53