<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] Remember that $\displaystyle \cos(\alpha) - \cos(\beta) = 2\sin\left(\frac{\alpha+\beta}{2}\right)\sin\left(\frac{\beta-\alpha}{2}\right)$. Hence we obtain
\begin{align*}
\cos(\theta) - \sin(\theta) = \cos(\theta) - \cos\left(\frac{\pi}{2}-\theta\right) = 2\sin\left(\frac{\pi}{4}\right)\sin\left(\frac{\pi}{4}-\theta\right) = \sqrt{2}\sin\left(\frac{\pi}{4}-\theta\right)
\end{align*}
Finally, we get:
\begin{align*}
\lim_{\theta\rightarrow\pi/4}\frac{\cos(\theta)-\sin(\theta)}{\theta-\displaystyle\frac{\pi}{4}} = \lim_{\theta\rightarrow\pi/4}\frac{\sqrt{2}\displaystyle\sin\left(\frac{\pi}{4}-\theta\right)}{\theta-\displaystyle\frac{\pi}{4}} = \lim_{x\rightarrow 0}\frac{-\sqrt{2}\sin(x)}{x} = -\sqrt{2}
\end{align*}
Where it has been used the well known result
$$\lim_{x\rightarrow 0}\frac{\sin(x)}{x} = 1$$
</s>