Find $\lim\limits_{x \to \infty} x\sin\frac{11}{x}$ Find $$\lim_{x \to \infty} x\sin\left(\frac{11}{x}\right)$$
We know $-1\le \sin \frac{11}{x} \le 1 $ 
Therefore, $x\rightarrow \infty $
And so limit of this function does not exist. 
Am I on the right track? Any help is much appreciated.
 A: $$\sin(11/x)\underset{(+\infty)}{\sim}11/x$$
What can you deduce ?
Note : What you have stated is good however with the product with $x$ you cannot conclude that it converges or diverges with what you wrote
A: Hint
$$\lim\limits_{x \to \infty} x\sin\left(\frac{11}{x}\right)=11\lim\limits_{x \to 0^+} \frac{\sin(11x)}{11x}$$
A: write $$11\frac{\sin(\frac{11}{x})}{\frac{11}{x}}$$ and the Limit is $$11$$
A: It is true that
$$-1\le \sin \frac{11}{x} \le 1$$
but since $x\to \infty$ we have that $$\sin \frac{11}{x}\to0$$
thus the limit is in the indeterminate form $0\cdot \infty$.
To solve we can set for example $y=\frac1x\to 0$ then
$$\lim_{x \to \infty} x\sin\left(\frac{11}{x}\right)=\lim_{y \to 0} \frac{\sin\left(11y\right)}{y}=11\cdot\lim_{y \to 0} \frac{\sin\left(11y\right)}{11y}=11$$
A: No the reasoning doesn't follow. If limit exists, then using your reasoning all we can say is it is between $-\infty$ and $\infty$. Make the change of variables $u=1/x$, and note that the limit is equivalent to
$$
\lim_{u\to 0^+}\frac{\sin 11u}{u}=11\lim_{u\to 0^+}\frac{\sin 11u}{11 u}
$$
and now use the well-known limit
$$
\lim_{x\to 0}\frac{\sin x}{x}=1
$$
A: Your conclusion is not directly correct, since you are neglecting the $x$ in the denominator inside the $\sin \frac{11}{x}$.
Write
$$ \lim_{x \rightarrow \infty} x \sin \frac{11}{x} = \lim_{x \rightarrow \infty} \frac {\sin \frac{11}{x}}{\frac{1}{x}}. $$
and L'Hôpital's rule is applicable.
