5
$\begingroup$

Let $f: [0, \infty) \longrightarrow \mathbb{R}$, where $\lim_{x \to \infty}f(x)$ exist, show that $\lim_{x \to \infty}f′(x)=0$

This fact is clearly intuitive to me but I could not write a rigorous demonstration of this.

$\endgroup$
  • 1
    $\begingroup$ This one is very famous and available on MSE, but as given by OP the statement is wrong. You need to add the assumption that $\lim_{x \to \infty}f'(x)$ exists. $\endgroup$ – Paramanand Singh Oct 27 '14 at 7:36
  • $\begingroup$ See the first result in this answer math.stackexchange.com/a/842293/72031 $\endgroup$ – Paramanand Singh Oct 27 '14 at 7:37
7
$\begingroup$

The fact is false. Here is a counterexample: enter image description here

You can smooth the corners so that $f'(x)$ is defined everywhere, but alternates between $+1$ and $-1$, so has no limit as $x\to \infty$. Meanwhile, $f(x)$ is squeezed between curves that approach zero, so $f(x)\to 0$.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ It's not true, but would it be true for non-oscillating functions? And if so, how would we then prove it? $\endgroup$ – Axoren Oct 27 '14 at 3:23
  • 1
    $\begingroup$ What does "non-oscillating" mean? You can make the red curve hug the top asymptote, then go at slope $-1$ to hug the bottom one, then up again. You can make this happen in a non-periodic way if you like, even at random, and the counterexample still stands. $\endgroup$ – vadim123 Oct 27 '14 at 3:28
  • $\begingroup$ @Axoren, it's not even true for monotonic functions. $\endgroup$ – Kaj Hansen Oct 27 '14 at 3:31
  • $\begingroup$ @KajHansen Could you give an example? It's late and I can't think of a monotonic function that it breaks for? $\endgroup$ – Axoren Oct 27 '14 at 3:41
  • 1
    $\begingroup$ @user67427, consider the image attached. We have a "step" that occurs at constant intervals, and that step gets smaller and smaller each time. However, at each step the derivative is $>>0$. From here, we can smooth out the steps to make the function differentiable. The resulting function is monotonic, limits to zero, and yet the derivative does not limit to zero. i.imgur.com/ePn2xxp.png $\endgroup$ – Kaj Hansen Oct 27 '14 at 4:15
6
$\begingroup$

This seems intuitive at first glance, but it's not true. Consider the following function:

$$f(x) = \frac{\sin(x^2)}{x+1}$$


To make this easier to see, consider instead the corresponding example for $g:(0, \infty) \rightarrow \mathbb{R}$:

$$g(x) = \frac{\sin(x^2)}{x}$$

And computing a derivative:

$$g'(x) = 2\cos(x^2) - \frac{\sin(x^2)}{x^2}$$

Certainly, $\displaystyle \lim_{x \rightarrow \infty} g(x) = 0$, but $\displaystyle \lim_{x \rightarrow \infty} g'(x)$ does not exist.

The above example was built out of this one so that it will be defined at $x = 0$.

| cite | improve this answer | |
$\endgroup$
1
$\begingroup$

First I think the necessary assumption is $f'(x)$ exist for all $x\in[0,\infty)$ otherwise the conterexamples are given in other answers. Here is another one, $f(x)=\frac{1}{x}$ when $x$ is rational and $0$ when $x$ is irrational, the function is discontinuous everywhere(therefore no derivative exist) but satisfies your criteria.

Assume $f'(x)$ exist everywhere, let $lim_{x\rightarrow\infty}f'(x)=a>0$. We can prove $lim_{x\rightarrow\infty}f(x)=\infty$, for given $\epsilon=\frac{a}{2}>0$, we can find $\delta>0$ such that $x>\delta$ implies $f'(x)>a-\epsilon=\frac{a}{2}>0$. We know continuous function($f'$ exist so f continuous) is integrable and we have $$f(\delta+\gamma)=f(\delta)+\int_{\delta}^{\delta+\gamma}f'(x)dx>f(\delta)+\frac{a\gamma}{2}$$ for any $\gamma>0$. This implies $f(x)$ can be arbitrary large for large enough $x$, and that's what $lim_{x\rightarrow\infty}f(x)=\infty$ means.

The prove of the cases $a<0$ or $a=\pm\infty$ are similar.

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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