# Proving the limit at $\infty$ of the derivative $f'$ is $0$ if it and the limit of the function $f$ exist. [duplicate]

Suppose that $f$ is differentiable for all $x$, and that $\lim_{x\to \infty} f(x)$ exists.

Prove that if $\lim_{x\to \infty} f′(x)$ exists, then $\lim_{x\to \infty} f′(x) = 0$, and also, give an example where $\lim_{x\to \infty} f′(x)$ does not exist.

I'm at a loss as to how to prove the first part, but for the second part, would a function such as $\sin(x)$ satisfy the problem?

## marked as duplicate by Hans Lundmark, user228113, user91500, John B, WatsonMar 26 '16 at 10:23

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

• $\sin$ is not an example for the second part, since $\lim_{x\to\infty} \sin x$ doesn't exist. – Daniel Fischer Dec 13 '14 at 23:57
• try something similar, like $\dfrac{\sin(x^2)}x$. From the graph it looks like the slope oscillates between $-1$ and $1$, and it ought to be easy to make this more formal. – Mirko Dec 14 '14 at 0:03
• @user48481MirkoSwirko and DanielFischer Thank you for your corrections. – angryavian Dec 14 '14 at 0:15
• For the second part: math.stackexchange.com/questions/162078/… – Hans Lundmark Mar 26 '16 at 9:36

## 2 Answers

Assuming both limits exist, apply L'Hospitals rule:

$$L= \lim_{x \rightarrow \infty}f(x) = \lim_{x \rightarrow \infty}\frac{e^xf(x)}{e^x}\\=\lim_{x \rightarrow \infty}[f(x) + f'(x)] = L + \lim_{x \rightarrow \infty} f'(x) ,$$

and conclude that $\lim_{x \rightarrow \infty} f'(x)= 0$.

Alternatively, by the MVT there is a point $\xi_x \in(x,x+1)$ such that

$$f(x+1)-f(x)=f'(\xi_x)$$

If $f'(x) \rightarrow L'$, then for every $\epsilon > 0$ there is a $K>0$ such that $|f'(x) - L'| < \epsilon$ when $x > K$. As $\xi_x > x > K$, it follows that $|f'(\xi_x) - L'| < \epsilon.$

Hence,

$$L'=\lim_{x \rightarrow \infty}f'(x)=\lim_{x \rightarrow \infty}f'(\xi_x)=\lim_{x \rightarrow \infty}[f(x+1)-f(x)]=0.$$

• I've seen this before, but I'm confused by the $e^x$. Can you please elaborate on the logic behind it? – user181928 Dec 14 '14 at 2:48
• That's Hardy's trick. However, I added a second proof which is more intuitive. The slope of the secant and the tangent must go to $0$ together. – RRL Dec 14 '14 at 3:12
• Ohhh well that makes a lot more sense, thanks for clarifying! – user181928 Dec 14 '14 at 3:43

If $\lim_{x\rightarrow\infty}f'(x)=c$ were some positive number, that would imply, for some $0<k<c$ and all large enough $x$ that $f'(x)>k$. Think about what this means intuitively and why this is inconsistent with $f$ converging.

• So I'm having a bit of trouble with this; would it mean that at some point $f'(x)$ would be less than k? Like I'm thinking of a function whose derivative is, for example, $\frac{2x-1}{x}$. Then choose k, so at some point we will have $f(x) \lt k$? Am I getting there, or am I not on the right track? – user181928 Dec 14 '14 at 1:20
• We're only interested in when $f'(x)>k$ - which must happen, by definition of a limit, for all $x>N$ for some $N$. So, if the derivative were $\frac{2x-1}x$, we might choose $k=1$ and $N=1$ - meaning that $f'(x)$ was always at least $1$ from there until infinity. – Milo Brandt Dec 14 '14 at 1:30
• Ok so I found something online: if $f+f′→L$ as $x→∞$ then $f→L$ and $f′→0$ because $lim_{x→∞}f(x) = lim_{x→∞}\frac{e^xf(x)}{e^x} = lim_{x→∞}\frac{e^x(f(x) + f'(x))}{e^x} = lim_{x→∞}(f(x) - f'(x))$ This would tell me that $f'(x)$ has to be zero since we said $f(x)$ is L. I don't quite understand this though; how did $e^x$ come into play? – user181928 Dec 14 '14 at 1:51
• @user I haven't the faintest idea - that doesn't look like it's effectively proving anything (and much less that it's proving what you're trying to). What I'm trying to get at is if $f'(x)>k$ for all $x>N$, then $f(x)>kx + f(N)$ for $x>N$ - that is, we can bound it below by a linear function. – Milo Brandt Dec 14 '14 at 2:04
• So I'm effectively going to try and do the same thing to bound it above, then sandwich it to $0$? Sorry if I'm jumping to assumptions, thanks for being very patient with me.. – user181928 Dec 14 '14 at 2:10