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?

Note that we have an application of the squeeze theorem here. We do not have L'Hopital's rule for the reasons mentioned above. However, we do have
$$
-\frac{x}{x^2+1}\leq \frac{x\sin(x)}{x^2+1}\leq\frac{x}{x^2+1}.
$$
using the bound $-1\leq\sin(x)\leq 1$. Since the limit of the upper and lower bounds go to $0$ as $x\to\infty$, the Squeeze theorem implies that 
$$
\lim_{x\to\infty}\frac{x\sin(x)}{x^2+1}=0.~_{\square}
$$
