Hi there,

I need an example of a function f for which lim x->infinity f(x) exists, but lim x-> inifinity f '(x) doesn't.

- Nov 15th 2009, 12:24 PMdgmath[SOLVED] limit of function exists, but limit of its derivative doesn't.
Hi there,

I need an example of a function f for which lim x->infinity f(x) exists, but lim x-> inifinity f '(x) doesn't. - Nov 15th 2009, 01:39 PMtonio
- Nov 15th 2009, 01:59 PMJhevon
- Nov 15th 2009, 02:18 PMdgmath
- Nov 15th 2009, 02:21 PMJhevon
- Nov 15th 2009, 02:41 PMJose27
Sorry, wrong argument.

Edit: What about $\displaystyle \frac{ \sin (x)}{ \sqrt{x} }$ ? - Nov 15th 2009, 02:43 PMJhevon
- Nov 15th 2009, 03:03 PMredsoxfan325
$\displaystyle f(x)=\frac{\sin(x^2)}{x}$ does it.

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

Clearly, $\displaystyle \lim_{x\to\infty}f(x)=0$, but $\displaystyle \lim_{x\to\infty}f'(x)=\lim_{x\to\infty}2\cos(x^2)$, which is undefined. - Nov 15th 2009, 03:06 PMJhevon
- Nov 15th 2009, 03:07 PMredsoxfan325
- Nov 15th 2009, 03:33 PMtonio
- Nov 15th 2009, 03:34 PMdgmath
umm, I might be acting stupid here, but I dont think that's quite right. because,

lim x-> inf sin(x^2)/x = (sin(x^2)/x^2) * (x^2/(x)) = lim x->inf 1 * x = infinity. Isn't that right? In that case, it wont work.

And if so, I think the actualy function that'd work is sinx/x^2 !!! because it will lead to limit 0. and derivative would be 2sinx/x - cosx which is DNE? - Nov 15th 2009, 03:40 PMJhevon
- Nov 15th 2009, 03:45 PMJhevon
**redsoxfan**is correct.

If $\displaystyle f(x) = \frac {\sin x^2}x$ (the infinite limit of which is zero--by the Squeeze theorem), then $\displaystyle \lim_{x \to \infty} f'(x) = \lim_{x \to \infty} \left( 2 \cos x^2 - \frac {\sin x^2}{x^2}\right) = \lim_{x \to \infty} 2 \cos x^2 - \lim_{x \to \infty}\frac {\sin x^2}{x^2}$ $\displaystyle = \lim_{x \to \infty} 2 \cos x^2 - 0 = \text{DNE}$

If $\displaystyle f(x) = \frac {\sin x}{x^2}$, then $\displaystyle f'(x) = \frac {x^2 \cos x - 2x \sin x}{x^4}$

and the infinite limit of both those functions is zero - Nov 15th 2009, 03:45 PMredsoxfan325
$\displaystyle \lim_{x\to\infty}\frac{-1}{x}\leq\lim_{x\to\infty}\frac{\sin(x^2)}{x}\leq\ lim_{x\to\infty}\frac{1}{x}$

So the Squeeze Theorem says $\displaystyle \lim_{x\to\infty}\frac{\sin(x^2)}{x}=0$

The derivative of $\displaystyle \frac{\sin x}{x^2}$ is $\displaystyle \frac{\cos x}{x^2}-\frac{2\sin x}{x^3}$, which clearly goes to $\displaystyle 0$ as $\displaystyle x\to\infty$.