Find the limit: $\lim_{x\to\infty}x\sin(\tan\frac1x)$ How do I find the limit:
$$\lim_{x\to\infty}x\sin(\tan\frac1x)$$
 A: Set $x=\frac{1}{t}$. Note that $$\lim_{x \to \infty} x \sin \left(\tan \frac{1}{x}\right)=\lim_{t \to 0^{+}} \frac{\sin (\tan t)}{t}$$
Now, note that $$\lim_{t \to 0^{+}} \frac{\sin (\tan t)}{t}=\lim_{t \to 0^{+}} \frac{\sin (\tan t)}{\tan t} \times \frac{\tan t}{t}=\lim_{t \to 0^{+}} \frac{\sin (\tan t)}{\tan t} \times \lim_{t \to 0^{+}}\frac{\tan t}{t}$$
Setting $\tan t =u$, the limit becomes $$\lim_{u \to 0^{+}} \frac{\sin u}{u} \times \lim_{t \to 0^{+}} \frac{\sin t}{t} \times \lim_{t \to 0^{+}} \frac{1}{\cos t}=1$$
As proven here. 
A: Rewrite the limit as 
\begin{align}
L:=\lim\limits_{x\rightarrow\infty}x\sin\left(\tan\frac{1}{x}\right)=\lim\limits_{x\rightarrow\infty}\frac{\sin\left(\tan\frac{1}{x}\right)}{\frac{1}{x}}
\end{align}
Since both the numerator and the denominator tend to $0$ in the limit, this is an indeterminate form and we can apply L'Hospital's Rule.
\begin{align}
L&=\lim\limits_{x\rightarrow\infty}\frac{\sin\left(\tan\frac{1}{x}\right)}{\frac{1}{x}}\\
&=\lim\limits_{x\rightarrow\infty}\frac{-\frac{1}{x^2}\sec^2\left(\frac{1}{x}\right)\cos\left(\tan\frac{1}{x}\right)}{-\frac{1}{x^2}}\\
&=\lim\limits_{x\rightarrow\infty}\sec^2\left(\frac{1}{x}\right)\cos\left(\tan\frac{1}{x}\right)\\
\end{align}
By continuity of all the functions involved, we can pull the limits into the functions. Since $\frac{1}{x}\rightarrow 0$ as $x\rightarrow\infty$, we have

$$\lim\limits_{x\rightarrow\infty}x\sin\left(\tan\frac{1}{x}\right) = 1$$

