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

Show 40 post(s) from this thread on one page
Page 1 of 2 12 Last
• Nov 15th 2009, 12:24 PM
dgmath
[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 PM
tonio
Quote:

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
• Nov 15th 2009, 01:59 PM
Jhevon
Quote:

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
• Nov 15th 2009, 02:18 PM
dgmath
Quote:

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.
• Nov 15th 2009, 02:21 PM
Jhevon
Quote:

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 :p) so i would keep barking up that tree.
• Nov 15th 2009, 02:41 PM
Jose27
Sorry, wrong argument.

Edit: What about $\frac{ \sin (x)}{ \sqrt{x} }$ ?
• Nov 15th 2009, 02:43 PM
Jhevon
Quote:

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...
• Nov 15th 2009, 03:03 PM
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.
• Nov 15th 2009, 03:06 PM
Jhevon
Quote:

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!
• Nov 15th 2009, 03:07 PM
redsoxfan325
Quote:

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.
• Nov 15th 2009, 03:33 PM
tonio
Quote:

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.(Headbang)

Tonio
• Nov 15th 2009, 03:34 PM
dgmath
Quote:

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?
• Nov 15th 2009, 03:40 PM
Jhevon
Quote:

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

Tonio

I don't think that would have worked either. squaring the argument seems to have been the trick with this sort of function.
• Nov 15th 2009, 03:45 PM
Jhevon
Quote:

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
• Nov 15th 2009, 03:45 PM
redsoxfan325
Quote:

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 $\lim_{x\to\infty}\frac{\sin(x^2)}{x}=0$

Quote:

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 $x\to\infty$.
Show 40 post(s) from this thread on one page
Page 1 of 2 12 Last