How can I calculate the following difficult limit: $\lim_{x\to\infty} (x + \sin x)\sin \frac1x$? How can I calculate the following limit?
$$\begin{equation*}
\lim_{x \rightarrow \infty}
(x + \sin x)\sin \frac{1}{x}
\end{equation*},$$
First, I multiplied and then distributed the limit then the limit of the first term was 1 but the second term was $\sin (x) \sin (1/x)$, I used the rule $\sin x \sin y = 1/2\{\cos(x-y) - \cos(x+y)\}$, but I got stucked, any help will be appreciated.
Thanks! 
 A: To find this limit, you're going to need to employ the following two well-known facts from the theory of limits:
$$
\lim_{\theta \rightarrow 0}\frac{\sin\theta}{\theta}=1
$$
and
$$
\lim_{\theta \rightarrow \infty}\frac{\sin\theta}{\theta}=0
$$
The proofs of those two statements are here and here respectively.
Notice that as $x$ approaches infinity, $\frac{1}{x}$ approaches $0$. So, $x\rightarrow \infty\implies\frac{1}{x}\rightarrow 0$. Also, don't forget that $x$ is the same thing as $\frac{1}{\frac{1}{x}}$ as long as $x$ does not equal zero (it doesn't in our case here):
$$
\lim_{x \rightarrow \infty}\left[(x + \sin x)\cdot\sin\frac{1}{x}\right]=
\lim_{x \rightarrow \infty}\left(x\cdot\sin\frac{1}{x}\right)
+\lim_{x \rightarrow \infty}\left(\sin x\cdot\sin\frac{1}{x}\right)=\\
\lim_{\frac{1}{x} \rightarrow 0}\frac{\sin\frac{1}{x}}{\frac{1}{x}}+
\lim_{x \rightarrow \infty}\left(\frac{x}{x}\cdot\sin x\cdot\sin\frac{1}{x}\right)=\\
1+\lim_{x \rightarrow \infty}\frac{\sin x}{x}\cdot
\lim_{\frac{1}{x}\rightarrow 0}\frac{\sin \frac{1}{x}}{\frac{1}{x}}=\\
1 + 0\cdot 1=1
$$
A: $$\lim_{x \rightarrow \infty}
(x + \sin x)\sin \frac{1}{x}=\lim_{x\rightarrow\infty}\frac{\sin\frac{1}{x}}{\frac{1}{x}}+0=1$$
