Find $\lim\limits_{x\to \infty} \frac{x\sin x}{1+x^2}$ 
$$\lim_{x\to \infty} \frac{x\sin x}{1+x^2}$$

Using L'hopital I get: 
$$\lim_{x\to \infty} \frac{x\cos x + \sin x}{2x}=\lim_{x \to \infty}\frac{\cos x}{2}$$
However, how is it possible to evaluate this limit?

Some answers are saying L'Hopital is not applicable here because the numerator does not $\to \pm \infty.$ Actually L'Hopital is valid if we assume only the denominator $\to \pm \infty:$ If $f,g$ are differentiable on $(a,\infty), \lim_{x\to \infty} g(x) = \pm \infty,$ and $\lim_{x\to\infty} \frac{f'(x)}{g'(x)} = L,$ then $\lim_{x\to\infty} \frac{f(x)}{g(x)} = L.$
For some reason this "better" L'Hopital is not as well known as the usual L'Hopital. It should be, because the proof is about as easy as the usual one, and the result resembles its cousin, the Stolz-Cesaro theorem for convergence of a sequence. (Recall that in SC, only the denominator sequence is assumed to $\to \pm \infty.$)
