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?
 A: You are trying this the hard way.  
Let $x = \frac\pi 4 -y$. Then
$$
\lim_{y\to 0}\frac{\cos(2(\frac\pi 4 -y))}{\sin(- y)}
=\lim_{y\to 0}\frac{\cos(\frac\pi 2-2y)}{-\sin y}
=-\lim_{y\to 0}\frac{\sin(2y)}{\sin y}
=-\lim_{y\to 0}\frac{2\sin y \cos y}{\sin y}
=-\lim_{y\to 0}2 \cos y=-2
$$
A: $$\cos(2x)=\sin\left(\frac{\pi}2-2x\right)\\\frac{\sin(\frac{\pi}2-2x)}{x-\frac{\pi}4}=\frac{2\sin(\frac{\pi}2-2x)}{2x-\frac{\pi}2}$$
Now it should be easy.
A: Let $u = x-\pi/4$ then you want $$\lim_{u\to 0} \frac{\cos(2(u + \pi/4))}{\sin u}=-\lim_{u \to 0} \frac{\sin 2u}{\sin u } = -\lim_{u \to 0} 2\cos u=-2$$
A: Let $f(x) = \cos (2x), g(x) = \sin (x-\pi/4).$ The expression equals
$$\frac{f(x) - f(\pi/4)}{g(x)-g(\pi/4)} = \dfrac{\dfrac{f(x) - f(\pi/4)}{x-\pi/4}}{\dfrac{g(x)-g(\pi/4)}{x-\pi/4}}.$$
By definition of the derivative and basic limit laws, the last expression $\to \dfrac{f'(\pi/4)}{g'(\pi/4)}$ as $x\to \pi/4,$  an easy computation. (And no, L'Hopital was nowhere used.) 
A: Just as an alternate approach:
I said: $\frac{cos(2\theta)}{sin\theta*cos(\frac{\pi}{4})-cos\theta*(\frac{\pi}{4})}$ =
$\frac{cos(2\theta)}{1/\sqrt(2)*sin\theta-cos\theta}$ Noting that $cos(2\theta)=cos^2\theta-sin^2\theta$ and $sin\theta-cos\theta=-(cos\theta-sin\theta)$
So, we have
$\frac{cos^2\theta-sin^2\theta}{-1/\sqrt(2)*(cos\theta-sin\theta)}$ = $-2*\frac{(cos\theta-sin\theta)*(\cos\theta+sin\theta)}{(cos\theta-sin\theta)}$=$-2*cos(\theta)+sin(\theta)$=-2 when you take the limit. Note that I just reciprocated the $-\frac{1}{\sqrt(2)}$
