4
$\begingroup$

Prove that if $\displaystyle \lim_{x \to \infty} f(x)$ and $\displaystyle \lim_{x \to \infty} f'(x)$ are both real numbers, then $\displaystyle \lim_{x \to \infty} f'(x) = 0$.

Attempt

Intuitively this makes sense to me. Take $y = x$. This slope is constant but it increases arbitrarily, and it seems that we can't make both the slope and value of $f(x)$ to be real numbers without "flattening" out the graph. I tried first saying $\displaystyle \lim_{x \to \infty} f(x) = a$ and $\displaystyle \lim_{x \to \infty} f'(x) = b$. Then we might be able to do something with the L'Hospital's rule.

$\endgroup$

marked as duplicate by Guy Fsone, Namaste calculus Jan 30 '18 at 0:59

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.

10
$\begingroup$

If $\lim_{x\to\infty}f(x)$ exists, then clearly

$$\lim_{x\to\infty}\left(1+{f(x)\over x}\right)=1$$

But ${x+f(x)\over x}$ qualifies for L'Hopitation, which gives

$$\lim_{x\to\infty}\left({x+f(x)\over x}\right)=\lim_{x\to\infty}\left({1+f'(x)\over 1}\right)=1+\lim_{x\to\infty}f'(x)$$

Hence $\lim_{x\to\infty}f'(x)=0$.

$\endgroup$
  • $\begingroup$ This is under assumption that $f(x)$ is a differentiable function on some open interval right? What if $f(x)$ is not differentiable? $\endgroup$ – user19405892 Mar 20 '16 at 0:16
  • 2
    $\begingroup$ @user19405892: Is $\lim_{x \to \infty}f'(x)$ meaningful if $f$ fails to be differentiable for all sufficiently large $x$? $\endgroup$ – Bungo Mar 20 '16 at 0:21
  • 2
    $\begingroup$ @user19405892, the assumption that $\lim_{x\to\infty}f'(x)$ exists implies that $f'(x)$ is defined for all sufficiently large $x$. $\endgroup$ – Barry Cipra Mar 20 '16 at 0:21
  • 2
    $\begingroup$ @Bungo, great minds think alike.... $\endgroup$ – Barry Cipra Mar 20 '16 at 0:22
  • 1
    $\begingroup$ @BarryCipra :-) Nice answer, btw (+1) $\endgroup$ – Bungo Mar 20 '16 at 0:23
4
$\begingroup$

$\displaystyle f(n+1)-f(n)=\frac{f(n+1)-f(n)}{n+1-n}=f'(c_n)$ where $c_n\in (n,n+1)$

Letting $n\to \infty$, $$0=\lim_\infty f'$$

$\endgroup$
  • $\begingroup$ How exactly does this arrive at the result? Do you assume continuity of f'? Is lim_n c_n = lim_n n? $\endgroup$ – BCLC Mar 20 '16 at 1:32
  • $\begingroup$ @BCLC No, I'm simply using the fact that $\lim_\infty f'$ exist. That implies $\lim_n f'(c_n)=\lim_x f'(x)$ $\endgroup$ – Gabriel Romon Mar 20 '16 at 1:38
  • $\begingroup$ LeGrandDODOM, how do you know that $\lim_{x \to \infty} f'(x)$ exists implies $\lim_{n \to \infty} f'(c_n) = \lim_{x \to \infty} f'(x)$? Oh, is it true for the similar reasons that reason that $\lim_{n \to \infty} f'(n) = \lim_{x \to \infty} f'(x)$ if $\lim_{x \to \infty} f'(x)$ exists is true? Or does it follow from the fact that $\lim_{n \to \infty} f'(n) = \lim_{x \to \infty} f'(x)$ if $\lim_{x \to \infty} f'(x)$ exists? $\endgroup$ – BCLC Mar 22 '16 at 13:49
  • 1
    $\begingroup$ @BCLC For the same reason that $\lim_n f(n)$ exists when $\lim_x f(x)$ exists $\endgroup$ – Gabriel Romon Mar 22 '16 at 14:53
1
$\begingroup$

I like Barry Cipra's answer. Here's another proof along the same lines: $$\lim_{x\to \infty} f(x) = \lim_{x\to \infty} {x f(x) \over x} = \lim_{x\to \infty} {x f'(x) + f(x) \over 1} = \lim_{x\to \infty} \big(x f'(x) + f(x)\big) $$ Now if $\lim_{x\to \infty} f'(x)>0$, $\lim_{x\to \infty} f(x)= +\infty$, and if $\lim_{x\to \infty}f'(x)<0$, $\lim_{x\to \infty} f(x) = -\infty$, both of which violate an assumption; hence $\lim_{x\to \infty} f'(x)=0$.

$\endgroup$
  • $\begingroup$ Do you assume $\lim_{x \to \infty} xf(x) = \infty$ ? $\endgroup$ – BCLC Mar 20 '16 at 0:22
  • $\begingroup$ If $\lim_{x \to \infty} xf'(x) = 0$, need we have that $\lim_{x \to \infty} f'(x) = 0$? $\endgroup$ – BCLC Mar 20 '16 at 0:24
  • 1
    $\begingroup$ @BCLC : If $\displaystyle\lim_{x\to\infty} f(x)=L$ is positive, then for large values of $x$, $f(x) > L/2$. Thus, $\displaystyle\lim_{x\to\infty} x f(x) \ge \lim_{x\to \infty} x \cdot L/2 = \infty$. Same logic if $L<0$. $\endgroup$ – Christopher Carl Heckman Mar 21 '16 at 7:11
  • 1
    $\begingroup$ @BCLC [second comment]: Remember that $\displaystyle\lim_{x\to\infty} f'(x)$ is a real number, so it's either positive, negative or zero. If it was positive, then we would have $\displaystyle\lim_{x\to\infty} xf'(x)=\infty$; if it were negative, we would have $\displaystyle\lim_{x\to\infty} xf'(x)=-\infty$. $\endgroup$ – Christopher Carl Heckman Mar 21 '16 at 7:14
  • $\begingroup$ Ah, I think I get it. Also, did you mean f' instead of f in the last part? $\endgroup$ – BCLC Mar 22 '16 at 13:56
1
$\begingroup$

Hint: $f(x)-f(x_0)=\int_{x_0}^x f'(t)\,dt$, especially for some $x_0$ sufficiently large and $x\ge x_0$.

$\endgroup$
  • $\begingroup$ Does your proof assume $$\int_{x_0}^{\infty} f'(t) dt$$ exists? $\endgroup$ – BCLC Mar 20 '16 at 0:20
  • 1
    $\begingroup$ @BCLC No, I think. It just uses the fact that $f'(t)> \frac b2$ for large $t$ if, say, $b>0$. $\endgroup$ – user228113 Mar 20 '16 at 0:29
  • 1
    $\begingroup$ @BCLC For the little it's worth, the hypothesis guarantees the existence of the improper Riemann integral even when $f'\cdot 1_{[0,\infty)}\notin L^1$. $\endgroup$ – user228113 Mar 20 '16 at 0:39
0
$\begingroup$

Employing the Mean value theorem, entails that, for each $x$ there exists $c_x\in (x,x+1)$ such that

$$f'(c_x)=f(x+1)-f(x)\to0$$

Hence $\lim_{x\to \infty}f'(x)=\lim_{x\to \infty}f'(c_x) =0$ since, $c_x\to\infty$ as $x\to\infty$ we have

$\endgroup$

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