The Limit of $e^x \sin(1/x)$ when $x$ approaches to infinity I want to compute $\lim_{x \to \infty} e^x \sin(1/x)$. Here what I did:
$\lim_{x \to \infty} e^x \sin(1/x) = \lim_{x \to \infty} \dfrac{\sin(1/x)}{e^{-x}}$. By using L'Hospital Rule I get $\lim_{x \to \infty} e^x \sin(1/x) = \lim_{x \to \infty} \dfrac{\cos(1/x) e^x}{x^2}$. 
What can I do right now? I know I can not use the product rule for limits as
$\lim_{x \to \infty} \dfrac{\cos(1/x) e^x}{x^2} = \lim_{x \to \infty} \cos(1/x) \cdot \lim_{x \to \infty} \dfrac{e^x}{x^2}$ 
since the second limit is infinity. Any help would be appreciated.
 A: Seems like you can use L'Hopital's rule. But here is a direct way to prove it :
If you know about Taylor expansion, you can use the fact that
\begin{align}
\sin(h) = h + o(h)
\end{align}
as $h$ goes to $0$, so that $\sin(1/x) = 1/x + o(1/x)$ as $x$ goes to $\infty$, and thus
\begin{align}
e^x \sin(1/x) = e^x/x + o(e^x/x)\end{align}
as $x$ goes to $\infty$, , thus the limit is $+\infty$
A: $e^{x} \sin (\frac  1 x)=(\frac {e^{x}} x) (x\sin (\frac  1 x) \to (\infty) (1)=\infty$.
[$\frac {e^{x}} x >\frac x  2 \to \infty$ and $\lim_{x \to \infty} x \sin (\frac  1 x)=\lim_{t \to 0+}\frac {\sin t } t =1$].
A: Another way, set $\frac{1}{x}=t$ and the limit $\lim_{t \to 0} \frac{\sin t }{t} = 1$:
$$
\lim_{t \to 0 } = e^{\frac{1}{t}}t \lim_{t \to 0}\frac{\sin t}{t} = \lim_{x \to \infty}\frac{e^x}{x} \times 1 = \infty
$$
A: You can indeed use the product rule for limits, since $1\cdot (+\infty)$ is not an indeterminate form. The result is $+\infty$. 
If you want a solution that does not use neither L'Hospital nor Taylor, you can just observe that
$$
e^x \sin (1/x) = \frac{e^x \sin(1/x)}{1/x}\frac{1}{x} = \frac{e^x}{x} \frac{\sin(1/x)}{1/x} \to (+\infty)\cdot1 = +\infty \quad \text{as } x \to +\infty 
$$
