inverse limits - Differentiability

Asked Dec 29 '19 at 11:36

Active Dec 29 '19 at 14:25

Viewed 37 times

suppose that $u$ is differentiable function and verifies: $$\lim_{x\to +\infty}u(x)=1$$ do we have $\lim_{x\to +\infty}u'(x)=0$?

My idea: we inserve limits

$$\lim_{x\to +\infty}u'(x)=lim_{x\to +\infty}\lim_{h\to 0}\frac{u(x+h)-u(x)}{h}=\lim_{h\to 0}lim_{x\to +\infty}\frac{u(x+h)-u(x)}{h}=\lim_{h\to 0}\frac{1-1}{h}=0$$

edited Dec 29 '19 at 11:39

asked Dec 29 '19 at 11:36

BrianTag

1,395
7
18

@mvpq yes! thank s for the remark – BrianTag Dec 29 '19 at 11:39
Why can you switch the limit operators here? – Clement Yung Dec 29 '19 at 11:42
3

Why do you believe the limit of $u'$ exists. Consider (vertical translations of) $u(x) = \sin(x^2)/x$. – Eric Towers Dec 29 '19 at 11:43
@EricTowers I supposed that $u$ is differentible. but I think this only works when $u'$ is continuous – BrianTag Dec 29 '19 at 11:45
$f(x) = \sin(x)$ is continuous (everywhere) and differentiable (everywhere) but neither $\lim_{x \rightarrow \infty} f(x)$ nor $\lim_{x \rightarrow \infty} f'(x)$ exist. – Eric Towers Dec 29 '19 at 11:47
@HansLundmark yes! Thank you for the suggestion – BrianTag Dec 29 '19 at 21:41

inverse limits - Differentiability

0 Answers0