Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Here is the homework problem I am stuck on:

Let $f$ be differentiable on $(0,\infty)$. If $\lim\limits_{x \to \infty} f'(x) = L$ exists in $\mathbb{R}$ and $\lim\limits_{n \to \infty} f(n) = A$ exists in $\mathbb{R}$, prove that $L = 0$.

From the given information, I know that we get to assume:

  1. $f$ is continuous at every point $s \in (0,\infty)$.
  2. We can now apply the MVT.

So far this is what I'm thinking: First, I think this describes a function which increases towards a horizontal asymptote. I have created a strictly increasing sequence $\{x_n\} = \{x_1, x_2, \dots, x_n \}$ to serve as the $x$ values in $(0, \infty)$. This gives me a sequence of intervals basically.

I can apply the MVT on each of these intervals, getting a sequence of $c_n \in (x_{n-1}, x_n)$.

But I don't see where to go from here. I am guessing I will need the squeeze theorem later, but there's a gap in between.

Perhaps I'm doing something wrong?

Thanks for your help.

share|cite|improve this question
Apply the MVT on each $[n,n+1]$: this yields $f(n+1)-f(n)=f'(c_n)$ for some $c_n\in (n,n+1)$. Let $n$ tend to $+\infty$, and observe that $c_n\longrightarrow +\infty$. – 1015 May 7 '13 at 23:03
if $L\neq0$, then $f'(x)>L-\epsilon$ (or $f'(x)<L+\epsilon$) for large $x$ and $f$ can be compared to a line explicitly. – yoyo May 8 '13 at 0:17
Thank you Julien! I was using $f(x_{n+1})$ and $f(x_n)$ instead of $f(n+1)$ and $f(n)$, which tripped me up. – FreshDresch May 8 '13 at 0:20

If we can show that $|L|\lt \epsilon$ for every $\epsilon \gt 0$ then we will have that $|L|\le 0\Rightarrow L=0$.

Let $\epsilon\gt 0$ be given. Since $$\lim_{x\to \infty} f^{\prime}(x)=L$$ we can find $M$ such that $x\ge M\Rightarrow |f^{\prime}(x) - L|\lt \frac{\epsilon}{2}$. Since $$\lim_{n\to\infty}f(n)=A$$ we can find $N$ such that $n\ge N\Rightarrow |f(n)-A|\lt \frac{\epsilon}{4}$.

Consider $m\gt max(M,N)$. The mean value theorem tells us we can find $c\in (m,m+2)$ such that $$f^{\prime}(c) = \frac{f(m+2) - f(m)}{2}$$ Then $|f^{\prime}(c)| = |\frac{f(m+2) - f(m)}{2}|\lt |f(m+2)-f(m)|\le |A+\frac{\epsilon}{4} - (A-\frac{\epsilon}{4})|=\frac{|\epsilon|}{2}=\frac{\epsilon}{2}$

Now, $|f^{\prime}(c)-L| = |L-f^{\prime}(c)|\lt\frac{\epsilon}{2}\Rightarrow -\frac{\epsilon}{2}\lt L-f^{\prime}(c)\lt \frac{\epsilon}{2}\Rightarrow f^{\prime}(c)-\frac{\epsilon}{2}\lt L\lt f^{\prime}(c)+\frac{\epsilon}{2}$.

But $-\frac{\epsilon}{2}\lt f^{\prime}(c)\lt \frac{\epsilon}{2}$ so $-\epsilon = \frac{-\epsilon}{2} + \frac{-\epsilon}{2}\lt f^{\prime}(c)+\frac{-\epsilon}{2}\lt L\lt f^{\prime}(c)+\frac{\epsilon}{2}\lt \frac{\epsilon}{2} + \frac{\epsilon}{2}=\epsilon$.

Therefore, $|L|\lt \epsilon$ and must be equal to $0.$

share|cite|improve this answer

Since this is a homework, I will only give an outline:

  1. As noted by Herberto, $f'(x)>K$ on some interval $[N,\infty)$, for some positive number $K$.

  2. Since $\lim_{n\to\infty}f(n)=A$, there exists integer $n>N$ such that $|f(n+1)-f(n)|$ is very small (small than $K$).

  3. Now use MVT to show that this is impossible.

share|cite|improve this answer

The proof can be done by contradiction.

Assume that $L\neq 0$, and without loss of generality you can assume that $L>0$ (otherwise use $g(x)=-f(x)$). Since $\lim_{x\to\infty}f(x)=L>0$, for small enough $\epsilon>0$ such that $L-\epsilon>0$, there exist $N>0$ such that for all $x>N$ $f'(x)>L-\epsilon>0$.

Now we have a differentiable function on an open interval $(N,\infty)$ such that $f'(x)>0$ we can conclude that $f$ is increasing, but this would imply that $\lim_{x\to\infty}f(x)=\infty$ obtaining a contradiction.

share|cite|improve this answer
Not every increasing function tends to infinity though. – TCL May 8 '13 at 4:19
It is true that not every increasing function tends to infinity, but in this case since $f'(x)>L-\epsilon>0$ $\epsilon$ can be chosen such that $L-\epsilon>L/2$ which will imply that $\lim_{x\to\infty}f'(x)>0$ thus avoiding horizontal asymptotes therefore we can safely conclude that $\lim_{x\to\infty}f(x)=\infty$ – Heberto del Rio May 30 '13 at 2:41

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.