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?
 A: $$\lim_{\theta \to \frac {\pi}{4}} \frac {\cos \theta - \sin \theta}{\theta - \frac {\pi}{4}} $$
Let $x:=\theta-\pi /4$, the limit you want to find equals to
$$\lim_{x \to 0} \frac {\cos (x+\pi /4) - \sin (x+\pi /4)}{x} $$
$$=\lim_{x \to 0} -\frac {\sqrt 2(\sin (x+\pi /4)\cos (\pi /4)-\sin (\pi /4)\cos (x+\pi /4))}{x} $$
As something we (intentionally) got in the bracket is the compound angle formula for $\sin$, we have the above limit to equal
$$\lim_{x \to 0} -\frac {\sqrt 2(\sin (x+\pi /4-\pi /4))}{x} $$
$$=\lim_{x \to 0} -\frac {\sqrt 2\sin{x}}{x} $$
$$=-\sqrt 2$$
A: Let $\theta-\frac{\pi}{4}=x$. Hence, $\theta=\frac{\pi}{4}+x$ and 
Hence, $$\lim_{\theta \to \dfrac {\pi}{4}} \dfrac {\cos \theta - \sin \theta}{\theta - \dfrac {\pi}{4}}=\lim_{x\rightarrow0}\frac{\cos\left(\frac{\pi}{4}+x\right)-\sin\left(\frac{\pi}{4}+x\right)}{x}=-\sqrt2\lim_{x\rightarrow0}\frac{\sin{x}}{x}=-\sqrt2.$$
A: 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
$$\lim_{x\rightarrow 0}\frac{\sin(x)}{x} = 1$$
A: 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. 
A: Add and subtract $\cos\left(\frac{\pi}{4}\right) = \sin\left(\frac{\pi}{4}\right)$ in the numerator to get:
$$\lim_{\theta \to \dfrac {\pi}{4}} \dfrac {\cos \theta - \sin \theta}{\theta - \dfrac {\pi}{4}} = \lim_{\theta \to \dfrac {\pi}{4}} \dfrac {\cos \theta - \cos\left(\frac{\pi}{4}\right) - \left(\sin \theta - \sin\left(\frac{\pi}{4}\right)\right)}{\theta - \dfrac {\pi}{4}} $$
$$= \lim_{\theta \to \dfrac {\pi}{4}} \dfrac {\cos \theta - \cos\left(\frac{\pi}{4}\right)}{\theta - \dfrac {\pi}{4}} - \lim_{\theta \to \dfrac {\pi}{4}} \dfrac {\sin \theta - \sin\left(\frac{\pi}{4}\right)}{\theta - \dfrac {\pi}{4}} = -\sin\left(\dfrac {\pi}{4}\right) - \cos\left(\dfrac {\pi}{4}\right) = -\sqrt{2}$$
where we used the definition for derivatives of sine and cosine.
A: Both numerator and denominator go to zero, so use L'hopital's.  Take derivative of numerator and denominator.  Get $(-\sin\theta-\cos\theta)/1$.  Take $\lim_\theta\rightarrow \frac {\pi}{4} $.  Get $(-\sqrt {2}/2-\sqrt {2}/2)/1=-\sqrt 2$.  
