# Thread: [SOLVED] limit of function exists, but limit of its derivative doesn't.

1. ## [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.

2. Originally Posted by dgmath
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.

$f(x)=\frac{\sin x}{x}$

Tonio

3. Originally Posted by tonio
$f(x)=\frac{\sin x}{x}$

Tonio
I had thought about that example the moment i saw this problem. But i believe that $\lim_{x \to \infty} f'(x) = 0$ in this case

4. Originally Posted by Jhevon
I had thought about that example the moment i saw this problem. But i believe that $\lim_{x \to \infty} f'(x) = 0$ in this case
you're right. the derivative of sinx/x would be (cosx.x - sinx)/x^2 . And since the denominator still has x^2, the limit will be zero instead of DNE.

5. Originally Posted by dgmath
you're right. the derivative of sinx/x would be (cosx.x - sinx)/x^2 . And since the denominator still has x^2, the limit will be zero instead of DNE.
hehe, yes, i know. my instincts tell me that we do want to use trig functions though (they have been wrong before ) so i would keep barking up that tree.

6. Sorry, wrong argument.

Edit: What about $\frac{ \sin (x)}{ \sqrt{x} }$ ?

7. Originally Posted by Jose27
Take $x=k\pi \ k \in \mathbb{Z}$ then $f(x)=0$, and for all other $x$ $f(x)\neq 0$ so $\lim_{x\rightarrow \infty } f(x)$ does not exist.
you are mistaken. the limit here is zero. one of many ways to see this is to apply the Squeeze theorem...

8. $f(x)=\frac{\sin(x^2)}{x}$ does it.

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

Clearly, $\lim_{x\to\infty}f(x)=0$, but $\lim_{x\to\infty}f'(x)=\lim_{x\to\infty}2\cos(x^2)$, which is undefined.

9. Originally Posted by redsoxfan325
$f(x)=\frac{\sin(x^2)}{x}$ does it.

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

Clearly, $\lim_{x\to\infty}f(x)=0$, but $\lim_{x\to\infty}f'(x)=\lim_{x\to\infty}2\cos(x^2)$, which is undefined.
Haha, there's the tweak we needed!

10. Originally Posted by Jose27
Sorry, wrong argument.

Edit: What about $\frac{ \sin (x)}{ \sqrt{x} }$ ?
The derivative is $\frac{\cos x}{\sqrt{x}}-\frac{\sin x}{2x^{3/2}}$ so the limit still approaches zero.

11. Originally Posted by Jhevon
I had thought about that example the moment i saw this problem. But i believe that $\lim_{x \to \infty} f'(x) = 0$ in this case

You are right, it does....forgot to square the denominator.

Tonio

12. Originally Posted by redsoxfan325
$f(x)=\frac{\sin(x^2)}{x}$ does it.

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

Clearly, $\lim_{x\to\infty}f(x)=0$, but $\lim_{x\to\infty}f'(x)=\lim_{x\to\infty}2\cos(x^2)$, which is undefined.
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?

13. Originally Posted by tonio
You are right, it does....forgot to square the denominator.

Tonio
I don't think that would have worked either. squaring the argument seems to have been the trick with this sort of function.

14. Originally Posted by dgmath
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?
redsoxfan is correct.

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

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

and the infinite limit of both those functions is zero

15. Originally Posted by dgmath
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. This is not right.

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?
$\lim_{x\to\infty}\frac{-1}{x}\leq\lim_{x\to\infty}\frac{\sin(x^2)}{x}\leq\ lim_{x\to\infty}\frac{1}{x}$

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

Originally Posted by tonio
You are right, it does....forgot to square the denominator.

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

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