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?
 A: Try evaluating
$$\lim_{x\rightarrow\infty}\frac{x}{1+x^2}\sin(x)$$
keeping in mind that $\sin$ is a bounded function..
A: L'Hopital's Rule doesn't apply here, since $\lim_{x\to\infty}x\sin x$ is undefined, not either $\infty$ or $-\infty$. Indeed, it oscillates with larger and larger "waves" as $x \to \infty$, and so in some sense tends to both $\infty$ and $-\infty$ at the same time. So the limitand is not an indeterminate form as $x \to \infty$.
Instead, here's a hint: What are the values of the limits $\lim_{x\to\infty}\frac{\pm x}{x^2+1}$? Can you relate this to your original limit?
A: You have $$\left\vert \frac{x \sin x}{1+x^2}\right\vert \le \frac{\vert x \vert}{1+x^2}$$ for all $x \in \mathbb{R}$.
A: You can't apply L'Hospital rule here: it supposes numerator and denominator both tend to $0$ or to $\infty$.
Unfortunately, the numerator has no limit as $x\to\infty$.
This is a fine example that L'Hospital's rule is dangerous. Although for some, it seems to be the alpha and omega of limits computation, when it works, it is logically equivalent to using Taylor's formula at order $1$, and very often, asymptotic calculus with equivalent function is much swifter:
$$x\sin x =O(x), \enspace 1+x^2\sim_\infty x^2,\enspace\text{hence}\quad \frac{x\sin x}{1+x^2}=\frac1{x^2}O(x)=O\Bigl(\frac1x\Bigr)\to 0.$$
A: 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 $\to \pm \infty.$)
A: Don't think L'Hospital applies here - the top limit does not exist. Hence, you are forced resort to a more fundamental approach like pointed out in other answers
A: 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 
$$
\lim_{x\to\infty}\frac{x\sin(x)}{x^2+1}=0.~_{\square}
$$
