<s>[INST] Find $\lim_{x\rightarrow \frac{\pi}{4}}\frac{\cos(2x)}{\sin(x-\frac{\pi}{4})}$ without L'Hopital I tried:
$$\lim_{x\rightarrow  \frac{\pi}{4}}\frac{\cos(2x)}{\sin(x-\frac{\pi}{4})} =
\frac{\frac{\cos(2x)}{x-\frac{\pi}{4}}}{\frac{\sin(x-\frac{\pi}{4})}{x-\frac{\pi}{4}}}$$  and $$\begin{align}\frac{\cos(2x)}{x-\frac{\pi}{4}} &=
\frac{\cos(2x)}{\frac{4x-\pi}{4}} \\&= 
\frac{4\cos(2x)}{4x-\pi} = \,\,???\end{align}$$
What do I do next?
 [/INST] You are trying this the hard way.  
Let $x = \frac\pi 4 -y$. Then
$$
\lim_{y\to 0}\frac{\cos(2(\frac\pi 4 -y))}{\sin(- y)}
=\lim_{y\to 0}\frac{\cos(\frac\pi 2-2y)}{-\sin y}
=-\lim_{y\to 0}\frac{\sin(2y)}{\sin y}
=-\lim_{y\to 0}\frac{2\sin y \cos y}{\sin y}
=-\lim_{y\to 0}2 \cos y=-2
$$
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