Q:

Find the limit of: $\lim_{x\to0}(\frac{x^2+2}{x^2})^{\sin(x)}$

I strugle finding the $\lim_{x\to0}(\frac{x^2+2}{x^2})^{\sin(x)}$. 
I wanted to say that: let $1+t = \frac{x^2+2}{x^2}$, which gives me $x^2=\frac{2}{t}$ and there I don't know how to go any further. Until now, all the problems of that sort were of degree 1, and you could simply put $t$ in, play a bit with the exponentials and then get a solution. If it would be $\cos(x)$, I could say that $\cos(-x) = \cos(x)$, but that's unfortunately not true for $\sin(x)$.
I then tried to say that $\sin(x)=\sqrt{1-\cos^2(x)}$, which gives me after a few operations:
$\lim_{t\to0}(1+t)^{\sqrt{\cos^2(\frac{\sqrt{2}}{\sqrt{t}}})})$, but I don't know how to continue, neither do I know whether it's correct.

A:

Hint
Start with $y=\left(\frac{x^2+2}{x^2}\right)^{\sin x}$, then 
\begin{align*}
\ln y & = \sin x \ln\left(\frac{x^2+2}{x^2}\right)\\
& = \sin x \left[\ln (x^2+2) -2\ln(x)\right]\\
& = \sin x \left[\ln (x^2+2)\right] -2\sin x \left[\ln(x)\right].
\end{align*}
as $x \to 0$, the first term goes to $0$. So you now have to focus on the second term and write it as 
$$2\frac{\sin x}{x}x \ln (x).$$
Now use the fact that $\frac{\sin x}{x} \to 1$ as $x \to 0$. The other expression $x \ln x$ is easy to deal with using L'Hospital.

