$f$ continuous, $a_nneq2,lim_{n to infty}a_n=2, b_n=frac{f(a_n)-f(2)}{a_n-2}, b_n$ converges $Rightarrow f$...
$begingroup$
So I've been struggling with this prove/disprove question:
Let $f:mathbb Rrightarrow mathbb R$ be a continuous function and let $a_nneq2, b_n $ be two sequences such that $lim_{n to infty}a_n=2$ and $b_n=frac{f(a_n)-f(2)}{a_n-2}.$ Prove or disprove that if $b_n$ converges, then $f$ is differentiable at $x=2$.
Since f is continuous, I realized that $lim_{n to infty} f(a_n)=f(2)$, but that is essentially it. I can't seem to figure out how to deal with the "$frac{0}{0}$" limit. It did occur to me, though, that this limit looks like a derivative, however I have no idea how to deal with it when it comes to sequences (Heine doesn't seem to help me much here).
Thank you very much and have a beautiful day.
calculus limits functions derivatives
asked Jan 2 at 14:30
Amit ZachAmit Zach
595
$endgroup$
|
show 2 more comments
$begingroup$
So I've been struggling with this prove/disprove question:
Let $f:mathbb Rrightarrow mathbb R$ be a continuous function and let $a_nneq2, b_n $ be two sequences such that $lim_{n to infty}a_n=2$ and $b_n=frac{f(a_n)-f(2)}{a_n-2}.$ Prove or disprove that if $b_n$ converges, then $f$ is differentiable at $x=2$.
Since f is continuous, I realized that $lim_{n to infty} f(a_n)=f(2)$, but that is essentially it. I can't seem to figure out how to deal with the "$frac{0}{0}$" limit. It did occur to me, though, that this limit looks like a derivative, however I have no idea how to deal with it when it comes to sequences (Heine doesn't seem to help me much here).
Thank you very much and have a beautiful day.
calculus limits functions derivatives
asked Jan 2 at 14:30
Amit ZachAmit Zach
595
$endgroup$
2
$begingroup$
Hint: $f$ does not have to be differentiable, look for a counterexample.
$endgroup$
– Wojowu
Jan 2 at 14:33
$begingroup$
@Wojowu Oh well that wasn't hard. I was so concentrated on proving the statement that I forgot that it might be wrong. Thanks!
$endgroup$
– Amit Zach
Jan 2 at 14:36
$begingroup$
@Wojowu If $f$ is differentiable for every $xneq 2$, can I say that $b_n$ is bounded?
$endgroup$
– Amit Zach
Jan 2 at 14:37
$begingroup$
No, it needn't be bounded. Unless you are still assuming $b_n$ is convergent, then of course it is bounded.
$endgroup$
– Wojowu
Jan 2 at 14:41
$begingroup$
@Wojowu Hmm, I see. It seems that I have an example for that, but the problem is that the $f$ that I chose would be differentiable for every real $x$. I can't find an example where $f$ is differentiable for every real $x$ apart from $2$. Could you help me with this one too? Much appriciated.
$endgroup$
– Amit Zach
Jan 2 at 14:46
|
show 2 more comments
$begingroup$
So I've been struggling with this prove/disprove question:
Let $f:mathbb Rrightarrow mathbb R$ be a continuous function and let $a_nneq2, b_n $ be two sequences such that $lim_{n to infty}a_n=2$ and $b_n=frac{f(a_n)-f(2)}{a_n-2}.$ Prove or disprove that if $b_n$ converges, then $f$ is differentiable at $x=2$.
Since f is continuous, I realized that $lim_{n to infty} f(a_n)=f(2)$, but that is essentially it. I can't seem to figure out how to deal with the "$frac{0}{0}$" limit. It did occur to me, though, that this limit looks like a derivative, however I have no idea how to deal with it when it comes to sequences (Heine doesn't seem to help me much here).
Thank you very much and have a beautiful day.
calculus limits functions derivatives
asked Jan 2 at 14:30
Amit ZachAmit Zach
595
$endgroup$
So I've been struggling with this prove/disprove question:
Let $f:mathbb Rrightarrow mathbb R$ be a continuous function and let $a_nneq2, b_n $ be two sequences such that $lim_{n to infty}a_n=2$ and $b_n=frac{f(a_n)-f(2)}{a_n-2}.$ Prove or disprove that if $b_n$ converges, then $f$ is differentiable at $x=2$.
Since f is continuous, I realized that $lim_{n to infty} f(a_n)=f(2)$, but that is essentially it. I can't seem to figure out how to deal with the "$frac{0}{0}$" limit. It did occur to me, though, that this limit looks like a derivative, however I have no idea how to deal with it when it comes to sequences (Heine doesn't seem to help me much here).
Thank you very much and have a beautiful day.
calculus limits functions derivatives
calculus limits functions derivatives
asked Jan 2 at 14:30
Amit ZachAmit Zach
595
asked Jan 2 at 14:30
Amit ZachAmit Zach
595
edited Jan 2 at 14:58
Amit Zach
asked Jan 2 at 14:30
Amit ZachAmit Zach
595
asked Jan 2 at 14:30
Amit ZachAmit Zach
595
asked Jan 2 at 14:30
Amit ZachAmit Zach
595
595
2
$begingroup$
Hint: $f$ does not have to be differentiable, look for a counterexample.
$endgroup$
– Wojowu
Jan 2 at 14:33
$begingroup$
@Wojowu Oh well that wasn't hard. I was so concentrated on proving the statement that I forgot that it might be wrong. Thanks!
$endgroup$
– Amit Zach
Jan 2 at 14:36
$begingroup$
@Wojowu If $f$ is differentiable for every $xneq 2$, can I say that $b_n$ is bounded?
$endgroup$
– Amit Zach
Jan 2 at 14:37
$begingroup$
No, it needn't be bounded. Unless you are still assuming $b_n$ is convergent, then of course it is bounded.
$endgroup$
– Wojowu
Jan 2 at 14:41
$begingroup$
@Wojowu Hmm, I see. It seems that I have an example for that, but the problem is that the $f$ that I chose would be differentiable for every real $x$. I can't find an example where $f$ is differentiable for every real $x$ apart from $2$. Could you help me with this one too? Much appriciated.
$endgroup$
– Amit Zach
Jan 2 at 14:46
|
show 2 more comments
2
$begingroup$
Hint: $f$ does not have to be differentiable, look for a counterexample.
$endgroup$
– Wojowu
Jan 2 at 14:33
$begingroup$
@Wojowu Oh well that wasn't hard. I was so concentrated on proving the statement that I forgot that it might be wrong. Thanks!
$endgroup$
– Amit Zach
Jan 2 at 14:36
$begingroup$
@Wojowu If $f$ is differentiable for every $xneq 2$, can I say that $b_n$ is bounded?
$endgroup$
– Amit Zach
Jan 2 at 14:37
$begingroup$
No, it needn't be bounded. Unless you are still assuming $b_n$ is convergent, then of course it is bounded.
$endgroup$
– Wojowu
Jan 2 at 14:41
$begingroup$
@Wojowu Hmm, I see. It seems that I have an example for that, but the problem is that the $f$ that I chose would be differentiable for every real $x$. I can't find an example where $f$ is differentiable for every real $x$ apart from $2$. Could you help me with this one too? Much appriciated.
$endgroup$
– Amit Zach
Jan 2 at 14:46
2
2
$begingroup$
Hint: $f$ does not have to be differentiable, look for a counterexample.
$endgroup$
– Wojowu
Jan 2 at 14:33
$begingroup$
Hint: $f$ does not have to be differentiable, look for a counterexample.
$endgroup$
– Wojowu
Jan 2 at 14:33
$begingroup$
@Wojowu Oh well that wasn't hard. I was so concentrated on proving the statement that I forgot that it might be wrong. Thanks!
$endgroup$
– Amit Zach
Jan 2 at 14:36
$begingroup$
@Wojowu Oh well that wasn't hard. I was so concentrated on proving the statement that I forgot that it might be wrong. Thanks!
$endgroup$
– Amit Zach
Jan 2 at 14:36
$begingroup$
@Wojowu If $f$ is differentiable for every $xneq 2$, can I say that $b_n$ is bounded?
$endgroup$
– Amit Zach
Jan 2 at 14:37
$begingroup$
@Wojowu If $f$ is differentiable for every $xneq 2$, can I say that $b_n$ is bounded?
$endgroup$
– Amit Zach
Jan 2 at 14:37
$begingroup$
No, it needn't be bounded. Unless you are still assuming $b_n$ is convergent, then of course it is bounded.
$endgroup$
– Wojowu
Jan 2 at 14:41
$begingroup$
No, it needn't be bounded. Unless you are still assuming $b_n$ is convergent, then of course it is bounded.
$endgroup$
– Wojowu
Jan 2 at 14:41
$begingroup$
@Wojowu Hmm, I see. It seems that I have an example for that, but the problem is that the $f$ that I chose would be differentiable for every real $x$. I can't find an example where $f$ is differentiable for every real $x$ apart from $2$. Could you help me with this one too? Much appriciated.
$endgroup$
– Amit Zach
Jan 2 at 14:46
$begingroup$
@Wojowu Hmm, I see. It seems that I have an example for that, but the problem is that the $f$ that I chose would be differentiable for every real $x$. I can't find an example where $f$ is differentiable for every real $x$ apart from $2$. Could you help me with this one too? Much appriciated.
$endgroup$
– Amit Zach
Jan 2 at 14:46
|
show 2 more comments
1 Answer
1
active
oldest
votes
$begingroup$
The statement is false.
For example if you let $f(x) = |x - 2|$ and $a_n = 2 + frac{1}{n}$ for each $nin mathbb{N}$, then $$lim_{ntoinfty} a_n = 2$$ and
$$lim_{ntoinfty} b_n = lim_{ntoinfty} frac{f(a_n) - f(2)}{a_n - 2} = lim_{ntoinfty} 1 = 1 $$
So $b_n$ converges, but $f$ is not differentiable at $2$.
answered Jan 2 at 14:50
M47145M47145
3,26331130
$endgroup$
2
$begingroup$
Thanks! This is exactly the counterexample I eventually used (although it was $|x-2|$ for me rather than $|x|-2$).
$endgroup$
– Amit Zach
Jan 2 at 14:53
$begingroup$
Oh yes, that is a typo. It should be $|x-2|$.
$endgroup$
– M47145
Jan 2 at 15:00
add a comment |
Your Answer
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1 Answer
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1 Answer
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$begingroup$
The statement is false.
For example if you let $f(x) = |x - 2|$ and $a_n = 2 + frac{1}{n}$ for each $nin mathbb{N}$, then $$lim_{ntoinfty} a_n = 2$$ and
$$lim_{ntoinfty} b_n = lim_{ntoinfty} frac{f(a_n) - f(2)}{a_n - 2} = lim_{ntoinfty} 1 = 1 $$
So $b_n$ converges, but $f$ is not differentiable at $2$.
answered Jan 2 at 14:50
M47145M47145
3,26331130
$endgroup$
2
$begingroup$
Thanks! This is exactly the counterexample I eventually used (although it was $|x-2|$ for me rather than $|x|-2$).
$endgroup$
– Amit Zach
Jan 2 at 14:53
$begingroup$
Oh yes, that is a typo. It should be $|x-2|$.
$endgroup$
– M47145
Jan 2 at 15:00
add a comment |
$begingroup$
The statement is false.
For example if you let $f(x) = |x - 2|$ and $a_n = 2 + frac{1}{n}$ for each $nin mathbb{N}$, then $$lim_{ntoinfty} a_n = 2$$ and
$$lim_{ntoinfty} b_n = lim_{ntoinfty} frac{f(a_n) - f(2)}{a_n - 2} = lim_{ntoinfty} 1 = 1 $$
So $b_n$ converges, but $f$ is not differentiable at $2$.
answered Jan 2 at 14:50
M47145M47145
3,26331130
$endgroup$
2
$begingroup$
Thanks! This is exactly the counterexample I eventually used (although it was $|x-2|$ for me rather than $|x|-2$).
$endgroup$
– Amit Zach
Jan 2 at 14:53
$begingroup$
Oh yes, that is a typo. It should be $|x-2|$.
$endgroup$
– M47145
Jan 2 at 15:00
add a comment |
$begingroup$
The statement is false.
For example if you let $f(x) = |x - 2|$ and $a_n = 2 + frac{1}{n}$ for each $nin mathbb{N}$, then $$lim_{ntoinfty} a_n = 2$$ and
$$lim_{ntoinfty} b_n = lim_{ntoinfty} frac{f(a_n) - f(2)}{a_n - 2} = lim_{ntoinfty} 1 = 1 $$
So $b_n$ converges, but $f$ is not differentiable at $2$.
answered Jan 2 at 14:50
M47145M47145
3,26331130
$endgroup$
The statement is false.
For example if you let $f(x) = |x - 2|$ and $a_n = 2 + frac{1}{n}$ for each $nin mathbb{N}$, then $$lim_{ntoinfty} a_n = 2$$ and
$$lim_{ntoinfty} b_n = lim_{ntoinfty} frac{f(a_n) - f(2)}{a_n - 2} = lim_{ntoinfty} 1 = 1 $$
So $b_n$ converges, but $f$ is not differentiable at $2$.
answered Jan 2 at 14:50
M47145M47145
3,26331130
edited Jan 2 at 15:01
answered Jan 2 at 14:50
M47145M47145
3,26331130
answered Jan 2 at 14:50
M47145M47145
3,26331130
answered Jan 2 at 14:50
M47145M47145
3,26331130
3,26331130
2
$begingroup$
Thanks! This is exactly the counterexample I eventually used (although it was $|x-2|$ for me rather than $|x|-2$).
$endgroup$
– Amit Zach
Jan 2 at 14:53
$begingroup$
Oh yes, that is a typo. It should be $|x-2|$.
$endgroup$
– M47145
Jan 2 at 15:00
add a comment |
2
$begingroup$
Thanks! This is exactly the counterexample I eventually used (although it was $|x-2|$ for me rather than $|x|-2$).
$endgroup$
– Amit Zach
Jan 2 at 14:53
$begingroup$
Oh yes, that is a typo. It should be $|x-2|$.
$endgroup$
– M47145
Jan 2 at 15:00
2
2
$begingroup$
Thanks! This is exactly the counterexample I eventually used (although it was $|x-2|$ for me rather than $|x|-2$).
$endgroup$
– Amit Zach
Jan 2 at 14:53
$begingroup$
Thanks! This is exactly the counterexample I eventually used (although it was $|x-2|$ for me rather than $|x|-2$).
$endgroup$
– Amit Zach
Jan 2 at 14:53
$begingroup$
Oh yes, that is a typo. It should be $|x-2|$.
$endgroup$
– M47145
Jan 2 at 15:00
$begingroup$
Oh yes, that is a typo. It should be $|x-2|$.
$endgroup$
– M47145
Jan 2 at 15:00
add a comment |
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2
$begingroup$
Hint: $f$ does not have to be differentiable, look for a counterexample.
$endgroup$
– Wojowu
Jan 2 at 14:33
$begingroup$
@Wojowu Oh well that wasn't hard. I was so concentrated on proving the statement that I forgot that it might be wrong. Thanks!
$endgroup$
– Amit Zach
Jan 2 at 14:36
$begingroup$
@Wojowu If $f$ is differentiable for every $xneq 2$, can I say that $b_n$ is bounded?
$endgroup$
– Amit Zach
Jan 2 at 14:37
$begingroup$
No, it needn't be bounded. Unless you are still assuming $b_n$ is convergent, then of course it is bounded.
$endgroup$
– Wojowu
Jan 2 at 14:41
$begingroup$
@Wojowu Hmm, I see. It seems that I have an example for that, but the problem is that the $f$ that I chose would be differentiable for every real $x$. I can't find an example where $f$ is differentiable for every real $x$ apart from $2$. Could you help me with this one too? Much appriciated.
$endgroup$
– Amit Zach
Jan 2 at 14:46