How do I evaluate this limit without using L'Hospital or Series expansion? I don't know how to do this limit.
$$\lim_{x\to\infty}\left(\frac1{x^2\sin^2\frac 1x}\right)^\frac 1{x\sin\frac 1x-1}$$
And here it is as an image, with bigger font:

I tried substituting $t=1/x$ and I got something better looking but it still didn't work. What trick do I need to use here?
 A: Considering $\frac{1}{x\sin\frac1x -1}\ln\left(\frac{1}{x^2\sin^2\frac1x}\right)$,
We substitute $t=1/x$
$\frac{1}{\frac{\sin t}{t}-1}\ln\left(\frac{1}{\frac{1}{t^2}\sin^2 t}\right)=\frac{t}{\sin t -t}\ln\left(\frac{t^2}{\sin^2 t}\right)$
However $\lim_{u \to 1}{\frac{\ln (u)}{u-1}}=1$
So we can equivalently consider
$\frac{t}{\sin t -t}\left(\frac{t^2}{\sin^2 t}-1\right)=\frac{t}{\sin t -t}\left(\frac{t^2-\sin^2 t}{\sin^2 t}\right)$
$=-\frac{t\left(\sin t+t\right)}{\sin^2 t}$
$=-\frac{t^2}{\sin^2 t}\left(\frac{\sin t}{t}+1\right)$
$\to -2$ as $t\to 0$
Hence the limit is $e^{-2}$
A: You need to use a couple of tricks. We are going to use $$\lim_{x\to\infty}\frac{1}{x\sin\frac 1x}=1$$ and $$\lim_{y\to\infty}\left(1+\frac 1y\right)^y=e$$
So we rewrite your original expression as:$$\begin{align}L=\lim_{x\to\infty}\left(\frac1{x^2\sin^2\frac 1x}\right)^\frac 1{x\sin\frac 1x-1}&=\lim_{x\to\infty}\left(x^2\sin^2\frac 1x\right)^\frac {-1}{x\sin\frac 1x-1}\\&=\lim_{x\to\infty}\left(1+x^2\sin^2\frac 1x-1\right)^{\frac {-1}{x\sin\frac 1x-1}\frac{x\sin\frac 1x+1}{x\sin\frac 1x+1}}\end{align}$$
Now use $1/y=x^2\sin^2\frac 1x-1$ and $\lim_{x\to\infty}(x\sin\frac1x)=1$ to get
$$L=e^{-2}$$
A: Your idea of making the substitution $x\mapsto1/x$ is fine.  We then wish to evaluate the limit as $x\to 0$ of $f(x)$ where $f(x)$ is given by
$$f(x)=\left(\left(\frac{x}{\sin(x)}\right)^{\frac{1}{\frac{\sin(x)}x-1}}\right)^2\tag1$$
Let $y=\frac{\sin(x)}{x}-1$ in $(1)$ so that $y\to 0$ as $x\to0$.   We now seek the limit 
$$\lim_{y\to0}\left(\left(\frac1{y+1}\right)^{\frac{1}{y}}\right)^2\tag2$$
Given the limit definition of $e$ as $e\equiv\lim_{y\to0}(1+y)^{1/y}$ we find immediately that the limit in $(2)$ is $\frac1{e^2}$ and hence 
$$\lim_{x\to 0}\left(\frac1{x^2\sin^2(1/x)}\right)^{\frac1{x\sin(1/x)-1}}=\frac1{e^2}$$
