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

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• Nov 15th 2009, 04:15 PM
tonio
Quote:

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

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 $\displaystyle x\to\infty$.

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

Tonio
• Nov 15th 2009, 04:18 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 $\displaystyle \frac{sin x}{x}$ I forgot to square the denominator when differentiating and thus got a wrong derivative.

Tonio

Oh, ok. no prob. :D
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