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
 A: Near $\pi$, you have made the denominator bigger, not smaller, so your inequality goes the other way, which does you no good. You can show, for instance using the mean value theorem on $\sin$ and then squaring the result, that $0 \leq \sin(x)^2 \leq (\pi-x)^2$. Plug that inequality in to see what you get.
A: Hint: I think you will be more comfortable if you make the substitution $\pi-x=t$. So $x=\pi-t$, and $\sin(\pi-t)=\sin t$. Now we are looking at behaviour near $t=0$, familiar territory.  
A: $|\pi -x|<\pi \implies \frac{1}{|\pi-x|}>\pi$.  So, we need a different approach.
Note that $\sin^2 x=\sin^2(\pi-x)$ and that 
$$\lim_{ \pi -x \to 0}\frac{\sin(\pi-x)}{\pi-x}=1.$$
Can you finish now?
A: 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*}
