prove that the limit of the given piecewise function does not exist, using the formal definition.
$begingroup$
i'm having problem with this exercise. Can somebody give me a hint ?
The function is
f(x) =
begin{cases}
x^{2}, & text{if x ≤ 2 } \
8 - 2x, & text{if 2 < x}
end{cases}
is given that the $limlimits_{x to 2} f(x)$ does not exist. but how i prove with the formal definition ?
i use the definition ∀ε > 0 ∃δ > 0 : | x ∋ (a,b), 0<|x-x'|<δ and |f(x) - L|<ε
and tried, first fixing ε = 0.1 , x'= 2 then applying |1.9 - 2|< δ => δ > 0.1 and |4.2 - 4| < ε => ε > 0.2 so by this result the limit exist ?!
Thanks.
edit: I misunderstood the concept of the existence of a limit and if the function can be derived. then yes the limit exist. But still in my professor problem are saying that the limit does not exist. Translating it is says literaly "prove by the formal definition why the lim does not exist". Probably he was playing with the class to see what we would answer.
calculus limits
asked Dec 5 '18 at 15:45
Jeremias JuniorJeremias Junior
154
$endgroup$
add a comment |
$begingroup$
i'm having problem with this exercise. Can somebody give me a hint ?
The function is
f(x) =
begin{cases}
x^{2}, & text{if x ≤ 2 } \
8 - 2x, & text{if 2 < x}
end{cases}
is given that the $limlimits_{x to 2} f(x)$ does not exist. but how i prove with the formal definition ?
i use the definition ∀ε > 0 ∃δ > 0 : | x ∋ (a,b), 0<|x-x'|<δ and |f(x) - L|<ε
and tried, first fixing ε = 0.1 , x'= 2 then applying |1.9 - 2|< δ => δ > 0.1 and |4.2 - 4| < ε => ε > 0.2 so by this result the limit exist ?!
Thanks.
edit: I misunderstood the concept of the existence of a limit and if the function can be derived. then yes the limit exist. But still in my professor problem are saying that the limit does not exist. Translating it is says literaly "prove by the formal definition why the lim does not exist". Probably he was playing with the class to see what we would answer.
calculus limits
asked Dec 5 '18 at 15:45
Jeremias JuniorJeremias Junior
154
$endgroup$
$begingroup$
Unfortunately, during translation something seems to have been lost. The limit exists at $x = 2$, and in fact this function is continuous everywhere, including at $2$.
$endgroup$
– астон вілла олоф мэллбэрг
Dec 5 '18 at 16:03
$begingroup$
you are right, i bet the professor was probably playing with us making this kind of affirmation in the problem. Just to see what we would answer.
$endgroup$
– Jeremias Junior
Dec 5 '18 at 16:05
$begingroup$
I see. Also, to close the question, either write an answer yourself and accept it after some time, or accept the answer below when you can (if you like it, but there is everything to like about it): this will automatically close the question.
$endgroup$
– астон вілла олоф мэллбэрг
Dec 5 '18 at 16:06
$begingroup$
did it, thanks :)
$endgroup$
– Jeremias Junior
Dec 5 '18 at 16:08
$begingroup$
You are welcome! +1 for the question, and providing attention to answers/comments.
$endgroup$
– астон вілла олоф мэллбэрг
Dec 5 '18 at 16:08
add a comment |
$begingroup$
i'm having problem with this exercise. Can somebody give me a hint ?
The function is
f(x) =
begin{cases}
x^{2}, & text{if x ≤ 2 } \
8 - 2x, & text{if 2 < x}
end{cases}
is given that the $limlimits_{x to 2} f(x)$ does not exist. but how i prove with the formal definition ?
i use the definition ∀ε > 0 ∃δ > 0 : | x ∋ (a,b), 0<|x-x'|<δ and |f(x) - L|<ε
and tried, first fixing ε = 0.1 , x'= 2 then applying |1.9 - 2|< δ => δ > 0.1 and |4.2 - 4| < ε => ε > 0.2 so by this result the limit exist ?!
Thanks.
edit: I misunderstood the concept of the existence of a limit and if the function can be derived. then yes the limit exist. But still in my professor problem are saying that the limit does not exist. Translating it is says literaly "prove by the formal definition why the lim does not exist". Probably he was playing with the class to see what we would answer.
calculus limits
asked Dec 5 '18 at 15:45
Jeremias JuniorJeremias Junior
154
$endgroup$
i'm having problem with this exercise. Can somebody give me a hint ?
The function is
f(x) =
begin{cases}
x^{2}, & text{if x ≤ 2 } \
8 - 2x, & text{if 2 < x}
end{cases}
is given that the $limlimits_{x to 2} f(x)$ does not exist. but how i prove with the formal definition ?
i use the definition ∀ε > 0 ∃δ > 0 : | x ∋ (a,b), 0<|x-x'|<δ and |f(x) - L|<ε
and tried, first fixing ε = 0.1 , x'= 2 then applying |1.9 - 2|< δ => δ > 0.1 and |4.2 - 4| < ε => ε > 0.2 so by this result the limit exist ?!
Thanks.
edit: I misunderstood the concept of the existence of a limit and if the function can be derived. then yes the limit exist. But still in my professor problem are saying that the limit does not exist. Translating it is says literaly "prove by the formal definition why the lim does not exist". Probably he was playing with the class to see what we would answer.
calculus limits
calculus limits
asked Dec 5 '18 at 15:45
Jeremias JuniorJeremias Junior
154
asked Dec 5 '18 at 15:45
Jeremias JuniorJeremias Junior
154
edited Dec 5 '18 at 16:11
Jeremias Junior
asked Dec 5 '18 at 15:45
Jeremias JuniorJeremias Junior
154
asked Dec 5 '18 at 15:45
Jeremias JuniorJeremias Junior
154
asked Dec 5 '18 at 15:45
Jeremias JuniorJeremias Junior
154
154
$begingroup$
Unfortunately, during translation something seems to have been lost. The limit exists at $x = 2$, and in fact this function is continuous everywhere, including at $2$.
$endgroup$
– астон вілла олоф мэллбэрг
Dec 5 '18 at 16:03
$begingroup$
you are right, i bet the professor was probably playing with us making this kind of affirmation in the problem. Just to see what we would answer.
$endgroup$
– Jeremias Junior
Dec 5 '18 at 16:05
$begingroup$
I see. Also, to close the question, either write an answer yourself and accept it after some time, or accept the answer below when you can (if you like it, but there is everything to like about it): this will automatically close the question.
$endgroup$
– астон вілла олоф мэллбэрг
Dec 5 '18 at 16:06
$begingroup$
did it, thanks :)
$endgroup$
– Jeremias Junior
Dec 5 '18 at 16:08
$begingroup$
You are welcome! +1 for the question, and providing attention to answers/comments.
$endgroup$
– астон вілла олоф мэллбэрг
Dec 5 '18 at 16:08
add a comment |
$begingroup$
Unfortunately, during translation something seems to have been lost. The limit exists at $x = 2$, and in fact this function is continuous everywhere, including at $2$.
$endgroup$
– астон вілла олоф мэллбэрг
Dec 5 '18 at 16:03
$begingroup$
you are right, i bet the professor was probably playing with us making this kind of affirmation in the problem. Just to see what we would answer.
$endgroup$
– Jeremias Junior
Dec 5 '18 at 16:05
$begingroup$
I see. Also, to close the question, either write an answer yourself and accept it after some time, or accept the answer below when you can (if you like it, but there is everything to like about it): this will automatically close the question.
$endgroup$
– астон вілла олоф мэллбэрг
Dec 5 '18 at 16:06
$begingroup$
did it, thanks :)
$endgroup$
– Jeremias Junior
Dec 5 '18 at 16:08
$begingroup$
You are welcome! +1 for the question, and providing attention to answers/comments.
$endgroup$
– астон вілла олоф мэллбэрг
Dec 5 '18 at 16:08
$begingroup$
Unfortunately, during translation something seems to have been lost. The limit exists at $x = 2$, and in fact this function is continuous everywhere, including at $2$.
$endgroup$
– астон вілла олоф мэллбэрг
Dec 5 '18 at 16:03
$begingroup$
Unfortunately, during translation something seems to have been lost. The limit exists at $x = 2$, and in fact this function is continuous everywhere, including at $2$.
$endgroup$
– астон вілла олоф мэллбэрг
Dec 5 '18 at 16:03
$begingroup$
you are right, i bet the professor was probably playing with us making this kind of affirmation in the problem. Just to see what we would answer.
$endgroup$
– Jeremias Junior
Dec 5 '18 at 16:05
$begingroup$
you are right, i bet the professor was probably playing with us making this kind of affirmation in the problem. Just to see what we would answer.
$endgroup$
– Jeremias Junior
Dec 5 '18 at 16:05
$begingroup$
I see. Also, to close the question, either write an answer yourself and accept it after some time, or accept the answer below when you can (if you like it, but there is everything to like about it): this will automatically close the question.
$endgroup$
– астон вілла олоф мэллбэрг
Dec 5 '18 at 16:06
$begingroup$
I see. Also, to close the question, either write an answer yourself and accept it after some time, or accept the answer below when you can (if you like it, but there is everything to like about it): this will automatically close the question.
$endgroup$
– астон вілла олоф мэллбэрг
Dec 5 '18 at 16:06
$begingroup$
did it, thanks :)
$endgroup$
– Jeremias Junior
Dec 5 '18 at 16:08
$begingroup$
did it, thanks :)
$endgroup$
– Jeremias Junior
Dec 5 '18 at 16:08
$begingroup$
You are welcome! +1 for the question, and providing attention to answers/comments.
$endgroup$
– астон вілла олоф мэллбэрг
Dec 5 '18 at 16:08
$begingroup$
You are welcome! +1 for the question, and providing attention to answers/comments.
$endgroup$
– астон вілла олоф мэллбэрг
Dec 5 '18 at 16:08
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Note that indeed
$$limlimits_{x to 2^+} f(x)=limlimits_{x to 2^-} f(x)=4$$
therefore the limit exists and you can proceed with the definition to prove that.
answered Dec 5 '18 at 15:48
gimusigimusi
92.9k84494
$endgroup$
$begingroup$
Oh i made a huge mistake! on my interpretation.
$endgroup$
– Jeremias Junior
Dec 5 '18 at 15:51
$begingroup$
translating it is says literaly "prove by the formal definition why the lim does not exist".
$endgroup$
– Jeremias Junior
Dec 5 '18 at 15:57
$begingroup$
@JeremiasJunior Yes but here the limit exists, there we can't prove that!
$endgroup$
– gimusi
Dec 5 '18 at 15:58
$begingroup$
for sure he playing with the class, by giving this kind of affirmation. Thanks again.
$endgroup$
– Jeremias Junior
Dec 5 '18 at 16:01
$begingroup$
Maybe it is a typo for the question or for the function. You can prove at first by definition that it exists and the taking $f(x)=8-3x$ for $x>2$ prove by definiiton that it doesn't exist.
$endgroup$
– gimusi
Dec 5 '18 at 16:02
|
show 2 more comments
Your Answer
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Note that indeed
$$limlimits_{x to 2^+} f(x)=limlimits_{x to 2^-} f(x)=4$$
therefore the limit exists and you can proceed with the definition to prove that.
answered Dec 5 '18 at 15:48
gimusigimusi
92.9k84494
$endgroup$
$begingroup$
Oh i made a huge mistake! on my interpretation.
$endgroup$
– Jeremias Junior
Dec 5 '18 at 15:51
$begingroup$
translating it is says literaly "prove by the formal definition why the lim does not exist".
$endgroup$
– Jeremias Junior
Dec 5 '18 at 15:57
$begingroup$
@JeremiasJunior Yes but here the limit exists, there we can't prove that!
$endgroup$
– gimusi
Dec 5 '18 at 15:58
$begingroup$
for sure he playing with the class, by giving this kind of affirmation. Thanks again.
$endgroup$
– Jeremias Junior
Dec 5 '18 at 16:01
$begingroup$
Maybe it is a typo for the question or for the function. You can prove at first by definition that it exists and the taking $f(x)=8-3x$ for $x>2$ prove by definiiton that it doesn't exist.
$endgroup$
– gimusi
Dec 5 '18 at 16:02
|
show 2 more comments
$begingroup$
Note that indeed
$$limlimits_{x to 2^+} f(x)=limlimits_{x to 2^-} f(x)=4$$
therefore the limit exists and you can proceed with the definition to prove that.
answered Dec 5 '18 at 15:48
gimusigimusi
92.9k84494
$endgroup$
$begingroup$
Oh i made a huge mistake! on my interpretation.
$endgroup$
– Jeremias Junior
Dec 5 '18 at 15:51
$begingroup$
translating it is says literaly "prove by the formal definition why the lim does not exist".
$endgroup$
– Jeremias Junior
Dec 5 '18 at 15:57
$begingroup$
@JeremiasJunior Yes but here the limit exists, there we can't prove that!
$endgroup$
– gimusi
Dec 5 '18 at 15:58
$begingroup$
for sure he playing with the class, by giving this kind of affirmation. Thanks again.
$endgroup$
– Jeremias Junior
Dec 5 '18 at 16:01
$begingroup$
Maybe it is a typo for the question or for the function. You can prove at first by definition that it exists and the taking $f(x)=8-3x$ for $x>2$ prove by definiiton that it doesn't exist.
$endgroup$
– gimusi
Dec 5 '18 at 16:02
|
show 2 more comments
$begingroup$
Note that indeed
$$limlimits_{x to 2^+} f(x)=limlimits_{x to 2^-} f(x)=4$$
therefore the limit exists and you can proceed with the definition to prove that.
answered Dec 5 '18 at 15:48
gimusigimusi
92.9k84494
$endgroup$
Note that indeed
$$limlimits_{x to 2^+} f(x)=limlimits_{x to 2^-} f(x)=4$$
therefore the limit exists and you can proceed with the definition to prove that.
answered Dec 5 '18 at 15:48
gimusigimusi
92.9k84494
answered Dec 5 '18 at 15:48
gimusigimusi
92.9k84494
answered Dec 5 '18 at 15:48
gimusigimusi
92.9k84494
answered Dec 5 '18 at 15:48
gimusigimusi
92.9k84494
92.9k84494
$begingroup$
Oh i made a huge mistake! on my interpretation.
$endgroup$
– Jeremias Junior
Dec 5 '18 at 15:51
$begingroup$
translating it is says literaly "prove by the formal definition why the lim does not exist".
$endgroup$
– Jeremias Junior
Dec 5 '18 at 15:57
$begingroup$
@JeremiasJunior Yes but here the limit exists, there we can't prove that!
$endgroup$
– gimusi
Dec 5 '18 at 15:58
$begingroup$
for sure he playing with the class, by giving this kind of affirmation. Thanks again.
$endgroup$
– Jeremias Junior
Dec 5 '18 at 16:01
$begingroup$
Maybe it is a typo for the question or for the function. You can prove at first by definition that it exists and the taking $f(x)=8-3x$ for $x>2$ prove by definiiton that it doesn't exist.
$endgroup$
– gimusi
Dec 5 '18 at 16:02
|
show 2 more comments
$begingroup$
Oh i made a huge mistake! on my interpretation.
$endgroup$
– Jeremias Junior
Dec 5 '18 at 15:51
$begingroup$
translating it is says literaly "prove by the formal definition why the lim does not exist".
$endgroup$
– Jeremias Junior
Dec 5 '18 at 15:57
$begingroup$
@JeremiasJunior Yes but here the limit exists, there we can't prove that!
$endgroup$
– gimusi
Dec 5 '18 at 15:58
$begingroup$
for sure he playing with the class, by giving this kind of affirmation. Thanks again.
$endgroup$
– Jeremias Junior
Dec 5 '18 at 16:01
$begingroup$
Maybe it is a typo for the question or for the function. You can prove at first by definition that it exists and the taking $f(x)=8-3x$ for $x>2$ prove by definiiton that it doesn't exist.
$endgroup$
– gimusi
Dec 5 '18 at 16:02
$begingroup$
Oh i made a huge mistake! on my interpretation.
$endgroup$
– Jeremias Junior
Dec 5 '18 at 15:51
$begingroup$
Oh i made a huge mistake! on my interpretation.
$endgroup$
– Jeremias Junior
Dec 5 '18 at 15:51
$begingroup$
translating it is says literaly "prove by the formal definition why the lim does not exist".
$endgroup$
– Jeremias Junior
Dec 5 '18 at 15:57
$begingroup$
translating it is says literaly "prove by the formal definition why the lim does not exist".
$endgroup$
– Jeremias Junior
Dec 5 '18 at 15:57
$begingroup$
@JeremiasJunior Yes but here the limit exists, there we can't prove that!
$endgroup$
– gimusi
Dec 5 '18 at 15:58
$begingroup$
@JeremiasJunior Yes but here the limit exists, there we can't prove that!
$endgroup$
– gimusi
Dec 5 '18 at 15:58
$begingroup$
for sure he playing with the class, by giving this kind of affirmation. Thanks again.
$endgroup$
– Jeremias Junior
Dec 5 '18 at 16:01
$begingroup$
for sure he playing with the class, by giving this kind of affirmation. Thanks again.
$endgroup$
– Jeremias Junior
Dec 5 '18 at 16:01
$begingroup$
Maybe it is a typo for the question or for the function. You can prove at first by definition that it exists and the taking $f(x)=8-3x$ for $x>2$ prove by definiiton that it doesn't exist.
$endgroup$
– gimusi
Dec 5 '18 at 16:02
$begingroup$
Maybe it is a typo for the question or for the function. You can prove at first by definition that it exists and the taking $f(x)=8-3x$ for $x>2$ prove by definiiton that it doesn't exist.
$endgroup$
– gimusi
Dec 5 '18 at 16:02
|
show 2 more comments
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$begingroup$
Unfortunately, during translation something seems to have been lost. The limit exists at $x = 2$, and in fact this function is continuous everywhere, including at $2$.
$endgroup$
– астон вілла олоф мэллбэрг
Dec 5 '18 at 16:03
$begingroup$
you are right, i bet the professor was probably playing with us making this kind of affirmation in the problem. Just to see what we would answer.
$endgroup$
– Jeremias Junior
Dec 5 '18 at 16:05
$begingroup$
I see. Also, to close the question, either write an answer yourself and accept it after some time, or accept the answer below when you can (if you like it, but there is everything to like about it): this will automatically close the question.
$endgroup$
– астон вілла олоф мэллбэрг
Dec 5 '18 at 16:06
$begingroup$
did it, thanks :)
$endgroup$
– Jeremias Junior
Dec 5 '18 at 16:08
$begingroup$
You are welcome! +1 for the question, and providing attention to answers/comments.
$endgroup$
– астон вілла олоф мэллбэрг
Dec 5 '18 at 16:08