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Ok so I have been working through Calculus by Spivak and stumbled upon a theorem which I found hard to prove ,and solution in answer book seems to be wrong.So I need you to help me verify my proof.

Theorem: If $\lim\limits_{x\to \infty}f(x)$ and $\lim\limits_{x\to \infty}f'(x)$ exist then $\lim\limits_{x\to \infty}f'(x)=0$

My proof:

Let $\lim\limits_{x\to \infty}f'(x)=l>0$.Then since limit is greater then 0,the derivative has to be greater then 0 on some interval.

Now I claim that there is some N such that $x>N \implies f'(x)>0$.If there is no such N then for $\epsilon<l$ there is no satisfying N and thus limit would not exist.

Now for some N we have that for every $x>N \implies f'(x)>0$ then since it is always greater then 0,it is also greater then some $m$ which is arbitrarily close to 0,thus we have that $f'(y)>m$ on interval [N,x] where x and N are as before explained.

Due to mean value theorem on interval [N,x] it holds that for some $y\in [N,x]$ $f'(y)=\frac{f(x)-f(N)}{x-N}>m \implies f(x)>f(N)+(x-N)m$ thus the function $f$ gets arbitrarily large since $(x-N)m$ gets arbitrarily large as $x\to\infty$

Thus it is proven that $\lim\limits_{x\to \infty}f(x)$ can not exist.

Proof of case when $\lim\limits_{x\to \infty}f'(x)=l<0$ is very similar

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This is essentially correct, though the part where you choose $m$ is a little sketchy. Instead you could write: Since $\lim_{x\to\infty}{f'(x)}=c>0$, then for any $\epsilon>0$ there exists an $N$ such that if $x>N$, then $|f'(x)-c|<\epsilon$, so in particular $f'(x)>c-\epsilon$. Then we choose $m=c-\epsilon$.

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    $\begingroup$ I though about my choice,but I could not come up with way to do that,but this is really nice choice,thanks for answer. $\endgroup$ Nov 29, 2014 at 17:02
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    $\begingroup$ @VanioBegic You're welcome. $\endgroup$ Nov 29, 2014 at 17:03

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