TITLE: Calculate trig limit of type $\frac{0}{0}$ without L'Hopital
QUESTION [7 upvotes]: I am trying to figure out the solution to this Limit without using L'Hopital. 
$$
\lim \limits_{x \to \pi} \frac {(\tan (4x))^2 } {(x - \pi )^2}
$$
Any help would be greatly appreciated.

REPLY [5 votes]: Not necessary, but I would let $x-\pi=t$. Note that $\tan(4x)=\tan(4t+4\pi)=\tan(4t)$. We want the limit of 
$$\frac{\tan(4t)}{t}\cdot \frac{\tan(4t)}{t}$$
as $t\to 0$.
Note that 
$$\frac{\tan(4t)}{t}=\frac{4}{\cos(4t)}\frac{\sin(4t)}{4t}.$$
Now can you finish?