Compute lim (x→pi/4) (sinx - cosx)/cos2x without using L'hopital rule

[hndalama]

Compute lim (x→pi/4) (sinx - cosx)/cos2x without using L'hopital rule

my attempt was to use an identity to change the cos2x but i found that the denominator will still equal 0.

 

[lookagain]

my attempt was to use an identity to change the cos2x but i found that the denominator will still equal 0.

Use this:

\(\displaystyle \displaystyle\lim_{x\to\ \frac{\pi}{4}}\dfrac{sin(x) \ - \ cos(x)}{cos^2(x) \ - \ sin^2(x)} \ = \)

\(\displaystyle -1*\displaystyle\lim_{x\to\ \frac{\pi}{4}}\dfrac{sin(x) \ - \ cos(x)}{ sin^2(x) \ - \ cos^2(x) } \ = \)

\(\displaystyle -1*\displaystyle\lim_{x\to\ \frac{\pi}{4}}\dfrac{sin(x) \ - \ cos(x)}{ [sin(x) \ - \ cos(x)][sin(x) \ + \ cos(x)] } \ = \)

 

[hndalama]

thank you