4

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.

user19405892
  • 15,592

5 Answers5

10

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

Barry Cipra
  • 79,832
4

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

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

Gabriel Romon
  • 35,428
  • 5
  • 65
  • 157
  • How exactly does this arrive at the result? Do you assume continuity of f'? Is lim_n c_n = lim_n n? – BCLC Mar 20 '16 at 01:32
  • @BCLC No, I'm simply using the fact that $\lim_\infty f'$ exist. That implies $\lim_n f'(c_n)=\lim_x f'(x)$ – Gabriel Romon Mar 20 '16 at 01:38
  • 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? – BCLC Mar 22 '16 at 13:49
  • 1
    @BCLC For the same reason that $\lim_n f(n)$ exists when $\lim_x f(x)$ exists – Gabriel Romon Mar 22 '16 at 14:53
1

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

  • Does your proof assume $$\int_{x_0}^{\infty} f'(t) dt$$ exists? – BCLC Mar 20 '16 at 00:20
  • 1
    @BCLC No, I think. It just uses the fact that $f'(t)> \frac b2$ for large $t$ if, say, $b>0$. –  Mar 20 '16 at 00:29
  • 1
    @BCLC For the little it's worth, the hypothesis guarantees the existence of the improper Riemann integral even when $f'\cdot 1_{[0,\infty)}\notin L^1$. –  Mar 20 '16 at 00:39
1

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

Christopher Carl Heckman
  • 4,458
  • Do you assume $\lim_{x \to \infty} xf(x) = \infty$ ? – BCLC Mar 20 '16 at 00:22
  • If $\lim_{x \to \infty} xf'(x) = 0$, need we have that $\lim_{x \to \infty} f'(x) = 0$? – BCLC Mar 20 '16 at 00:24
  • 1
    @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 $L<0$. – Christopher Carl Heckman Mar 21 '16 at 07:11
  • 1
    @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 $\displaystyle\lim_{x\to\infty} xf'(x)=-\infty$. – Christopher Carl Heckman Mar 21 '16 at 07:14
  • Ah, I think I get it. Also, did you mean f' instead of f in the last part? – BCLC Mar 22 '16 at 13:56
0

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

Guy Fsone
  • 23,903