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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.

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    $\begingroup$ 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 $c_n\longrightarrow +\infty$. $\endgroup$
    – Julien
    May 7, 2013 at 23:03
  • $\begingroup$ 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. $\endgroup$
    – yoyo
    May 8, 2013 at 0:17
  • $\begingroup$ 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. $\endgroup$ May 8, 2013 at 0:20

3 Answers 3

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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 $0.$

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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.

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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.

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  • $\begingroup$ Not every increasing function tends to infinity though. $\endgroup$
    – TCL
    May 8, 2013 at 4:19
  • $\begingroup$ 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 $\lim_{x\to\infty}f(x)=\infty$ $\endgroup$ May 30, 2013 at 2:41

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