Question

Lim x tends to π/4 (4√2-(cos x + sin x)^5)/1- sin2x

Answer 1

limits:  
Apply L'Hopital's rule, as the numerator and denominator are approaching 0 at the limit x -> π/4.  So differentiate the numerator and denominator wrt x.

$L= \lim_{x \to \frac{\pi}{4}} F(x)\\\\F(x)= \frac{4\sqrt{2}-(cosx+sinx)^5}{1-sin2x}\\\\Limit=\frac{-5(cosx+sinx)^4*(cosx-sinx)}{-2cos2x}\\\\=\frac{-5(cosx+sinx)^4*(cosx-sinx)}{-2(cosx+sinx)(cosx-sinx)}\\\\=\frac{5}{2}(cosx+sinx)^3\\\\=\frac{5}{2}(cos\frac{\pi}{4}+sin\frac{\pi}{4})^3\\\\=5\sqrt{2}$

so answer = 5 √2

Answer 2

Here's your answer ———