Q:

Finding the value of the limit of $e^{2(g(x)+1)}$ for $x\to 0$

$\lim\limits_{x\to0} e^{2(g(x)+1)}$ if
$g(x)=
\begin{cases}
x^2\sin\left(\frac1x\right),&x\ne0\\
2,&x=0
\end{cases}$
Lost at how to begin solving this. Please give some advice, Thank you.
I am solving for a continuity. So my reasoning here is:

I guess the squeeze theorem for x^2sin(1/x), and it equals 0. Which confuses me because x is not supposed to equal 0 
The second term I assume that I would take a number very close to 0 like 0.0001. So then a  very small number squared times sin (1/x) will go close to infinity. ( reasoning 1 divided by a really small number would make a very large number.)

I am thrown off by $e^{2(g(x)+1)}$ fo $x\to 0$. I'm not getting how to use e as a base number

A:

Since $\left|x^2\sin\left(\frac{1}{x}\right)\right| \le |x^2|$ you have that 
$x^2\sin\left(\frac{1}{x}\right) \to 0$ as $x \to 0$. 
Then $2(g(x)+1) \to 2$ as $x \to 0, x \ne 0$.

