<s>[INST] Evaluate: $\lim_{\theta \to \frac {\pi}{4}} \dfrac {\cos \theta - \sin \theta}{\theta - \frac {\pi}{4}}$ 
Evaluate: $\lim_{\theta \to \frac {\pi}{4}} \dfrac {\cos \theta - \sin \theta}{\theta - \dfrac {\pi}{4}}$.

My Attempt:
\begin{align}
\lim_{\theta \to \frac {\pi}{4}} \dfrac {\cos \theta - \sin \theta }{\theta - \dfrac {\pi}{4}}
&=\lim_{\theta \to \frac {\pi}{4}} \dfrac {\cos \theta - \cos \dfrac {\pi}{4} + \sin \dfrac {\pi}{4} - \sin \theta}{\theta - \dfrac {\pi}{4}}
\\
&=\lim_{\theta \to \frac {\pi}{4}} \dfrac {2\sin \dfrac {\pi-4\theta }{8}\cos \dfrac {\pi+4\theta}{8} - 2\sin \dfrac {4\theta + \pi}{8}\sin \dfrac {4\theta -\pi}{8}}{\theta - \dfrac {\pi}{4}}.
\end{align}
How do I proceed?
 [/INST] Note: $cos(π/4) = sin(π/4) = 1/√2$.
$\frac{cos(\theta) - sin(\theta)}{\theta - π/4} =$
√2 $\frac{sin(π/4) cos(\theta) - cos(π/4)sin(\theta)}{\theta - π/4} = $
√2$ \frac{sin(π/4 - \theta)}{\theta - π/4} = $ 
-√2$ \frac{sin(x)}{x}$ , where  x: = $\theta - π/4$.
$\lim_{x \rightarrow 0}$ (-√2 $\frac{sin(x)}{x}) =$ -√2. 
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