How do you find the limit as x approaches pi/4 of [sin(x) - cos(x)] / cos(2x)?

Apr 16, 2015

$\frac{\sqrt{2}}{- 2}$

This is an indeterminate form of the type $\frac{0}{0}$, hence L'Hopital's rule would apply and limit can be evaluated by differentiating numerator and denominator and then applying the limit. Accordingly,

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

= Lim$x \to \frac{\pi}{4}$ $\frac{\cos x + \sin x}{- 2 \sin 2 x}$

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

Apr 16, 2015

If you want to evaluate the limit without l'Hopital's Rule (find the limit "algebraically"), do this:

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

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

$= {\lim}_{x \rightarrow \frac{\pi}{4}} \frac{- 1}{\cos x + \sin x} = \frac{- 1}{\frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}}} = \frac{- 1}{\frac{2}{\sqrt{2}}} = - \frac{\sqrt{2}}{2}$