# How do you find \lim _ { x \rightarrow - \frac { x } { 4} } \frac { \cos 2x } { \sin x + \cos x }?

Jan 15, 2018

${\lim}_{x \to - \frac{\pi}{4}} \cos \frac{2 x}{\sin \left(x\right) + \cos \left(x\right)} = \sqrt{2}$

#### Explanation:

First investigate: Substitute $- \frac{\pi}{4}$ for $x$

Since:

color(blue)(sin(-pi/4)=-sqrt2/2

color(blue)(cos(-pi/4)=sqrt2/2

${\lim}_{x \to - \frac{\pi}{4}} \cos \frac{2 x}{\sin \left(x\right) + \cos \left(x\right)} = \frac{0}{- \frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2}} = \frac{0}{0}$

We end up with the indeterminate form $\frac{0}{0}$ which implies we should use L'Hospitals Rule

This rule states:

If, ${\lim}_{x \to a} f \frac{x}{g} \left(x\right) = \frac{0}{0}$

Then, ${\lim}_{x \to a} f \frac{x}{g} \left(x\right) = {\lim}_{x \to a} \frac{f ' \left(x\right)}{g ' \left(x\right)}$

So,

lim_(x->-pi/4)cos(2x)/(sin(x)+cos(x))=lim_(x->-pi/4)(d/dx[cos(2x)])/(d/dx[(sin(x)+cos(x)]

$= {\lim}_{x \to - \frac{\pi}{4}} \frac{- 2 \sin \left(2 x\right)}{\cos \left(x\right) - \sin \left(x\right)}$

Substitute in $- \frac{\pi}{4}$ for $x$:

$= {\lim}_{x \to - \frac{\pi}{4}} \frac{- 2 \sin \left(2 \left(\textcolor{red}{- \frac{\pi}{4}}\right)\right)}{\cos \left(\textcolor{red}{- \frac{\pi}{4}}\right) - \sin \left(\textcolor{red}{- \frac{\pi}{4}}\right)} = \frac{- 2 \cdot - 1}{\frac{\sqrt{2}}{2} - \left(- \frac{\sqrt{2}}{2}\right)}$

$= \frac{2}{\frac{2 \sqrt{2}}{2}} = \frac{2}{\sqrt{2}}$

Simplify by applying the exponent rule: ${x}^{a} / {x}^{b} = {x}^{a - b}$

$\frac{2}{\sqrt{2}} = {2}^{1 - \frac{1}{2}} = {2}^{\frac{1}{2}} = \sqrt{2}$

Jan 15, 2018

Use $\cos 2 x = {\cos}^{2} x - {\sin}^{2} x$, then factor and reduce.

#### Explanation:

${\lim}_{x \rightarrow - \frac{\pi}{4}} \frac{\cos 2 x}{\sin x + \cos x} = {\lim}_{x \rightarrow - \frac{\pi}{4}} \frac{{\cos}^{2} x - {\sin}^{2} x}{\cos x + \sin x}$

$= {\lim}_{x \rightarrow - \frac{\pi}{4}} \frac{\left(\cos x - \sin x\right) \left(\cos x + \sin x\right)}{\cos x + \sin x}$

 = lim_(xrarr-pi/4)(cosx-sinx))

$= \cos \left(- \frac{\pi}{4}\right) - \sin \left(- \frac{\pi}{4}\right)$

$= \frac{\sqrt{2}}{2} - \left(- \frac{\sqrt{2}}{2}\right) = \sqrt{2}$