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.
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 $\displaystyle f(x) = \frac {\sin x^2}x$ (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
$\displaystyle \lim_{x\to\infty}\frac{-1}{x}\leq\lim_{x\to\infty}\frac{\sin(x^2)}{x}\leq\ lim_{x\to\infty}\frac{1}{x}$
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 $\displaystyle x\to\infty$.