How do you compute the limit of (sinx-cosx)/cos(2x) as x->pi/4?

Nov 19, 2016

Multiply by $\frac{\sin \left(x\right) + \cos \left(x\right)}{\sin \left(x\right) + \cos \left(x\right)}$.

$\frac{\sin \left(x\right) - \cos \left(x\right)}{\cos \left(2 x\right)} \cdot \frac{\sin \left(x\right) + \cos \left(x\right)}{\sin \left(x\right) + \cos \left(x\right)} = \frac{{\sin}^{2} \left(x\right) - {\cos}^{2} \left(x\right)}{\cos \left(2 x\right) \left(\sin \left(x\right) + \cos \left(x\right)\right)}$

Use the double angle identity for cosine: ${\cos}^{2} \left(x\right) - {\sin}^{2} \left(x\right) \equiv \cos \left(2 x\right)$

$\frac{{\sin}^{2} \left(x\right) - {\cos}^{2} \left(x\right)}{\cos \left(2 x\right) \left(\sin \left(x\right) + \cos \left(x\right)\right)} = \frac{- \cos \left(2 x\right)}{\cos \left(2 x\right) \left(\sin \left(x\right) + \cos \left(x\right)\right)}$

Now to evaluate the limit

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

$= - \frac{1}{\sin \left(\frac{\pi}{4}\right) + \cos \left(\frac{\pi}{4}\right)}$

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

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