limit of sin function as it approches $\pi$ In my assignment I have to find the Classification of discontinuities of the following function:
$$f(x)=\frac{\sin^2(x)}{x|x(\pi-x)|}$$
I wanted to look what happens with the value $x=\pi$ because the function doesn't exist in that value. 
I have to check if some  $L \in \Bbb R$ exists such that $$\lim  _{x \to \pi} f(x)=L $$ 
I didn't have much success in making the argument simpler, so I thought to make a little 'trick', and I know it works for sequences. I am not sure if it's "legal" to do in function. Here it is:
$$\frac{\sin^2(x)}{x|x(\pi-x)|}  < \frac{\sin^2(x)}{x|x(\pi )|}$$
Now find the limit for the "bigger" function:
$$\lim _{x \to \pi} \frac {sin^2(x)}{x|x(\pi )|} = \frac{0}{\pi}=0$$
Is my solution "legal" or valid?
Thanks,
Alan

Since $\sin x=\sin (\pi -x)$ then
\begin{eqnarray*}
\lim_{x\rightarrow \pi }\frac{\sin ^{2}(x)}{x\left\vert x(\pi -x)\right\vert 
} &=&\lim_{x\rightarrow \pi }\left( \frac{\sin (\pi -x)}{(\pi -x)}\right)
^{2}\frac{\left\vert \pi -x\right\vert }{x\left\vert x\right\vert } \\
&=&\lim_{x\rightarrow \pi }\left( \frac{\sin (\pi -x)}{(\pi -x)}\right)
^{2}\cdot \lim_{x\rightarrow \pi }\frac{\left\vert \pi -x\right\vert }{%
x\left\vert x\right\vert } \\
&=&1^{2}\cdot \frac{\left\vert \pi -\pi \right\vert }{\pi \left\vert \pi
\right\vert } \\
&=&1\cdot \frac{0}{\pi ^{2}}=0.
\end{eqnarray*}
I have used the standard limit
\begin{equation*}
\lim_{u\rightarrow 0}\frac{\sin (u)}{(u)}=1.
\end{equation*}
