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

Show 40 post(s) from this thread on one page
Page 2 of 2 First 12
• Nov 15th 2009, 05:15 PM
tonio
Quote:

Originally Posted by redsoxfan325
$\lim_{x\to\infty}\frac{-1}{x}\leq\lim_{x\to\infty}\frac{\sin(x^2)}{x}\leq\ lim_{x\to\infty}\frac{1}{x}$

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

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 $x\to\infty$.

This is a mess and everyone understands what she/he wants (Rock) : I meant that in my example $\frac{sin x}{x}$ I forgot to square the denominator when differentiating and thus got a wrong derivative.

Tonio
• Nov 15th 2009, 05:18 PM
Jhevon
Quote:

Originally Posted by tonio
This is a mess and everyone understands what she/he wants (Rock) : I meant that in my example $\frac{sin x}{x}$ I forgot to square the denominator when differentiating and thus got a wrong derivative.

Tonio

Oh, ok. no prob. :D
Show 40 post(s) from this thread on one page
Page 2 of 2 First 12