Evaluate this limit without using L' hospital rule

$\begin{aligned} L & = \lim_{x \to \frac \pi 4} \frac {4\sqrt 2-(\sin x+\cos x)^5}{1-\sin 2x} \\ & = \lim_{x \to \frac \pi 4} \frac {4\sqrt 2- 4\sqrt 2 \sin^5 \left(x+\frac \pi 4\right)}{1-\sin 2x} & \small \color{#3D99F6} \text{Let }u = x + \frac \pi 4 \\ & = \lim_{u \to \frac \pi 2} \frac {4\sqrt 2 (1 - \sin^5 u)}{1+\cos 2u} \\ & = \lim_{u \to \frac \pi 2} \frac {2\sqrt 2 (1 - \sin^5 u)}{\cos^2 u} & \small \color{#3D99F6} \text{A 0/0 case, L'Hôpital's rule applies} \\ & = \lim_{u \to \frac \pi 2} \frac {-10\sqrt 2 \sin^4 u \cos x}{-2\cos u\sin u} & \small \color{#3D99F6} \text{Differentiate up and down w.r.t. }u \\ & = \lim_{u \to \frac \pi 2} 5\sqrt 2 \sin^3 u \\ & = \boxed {5\sqrt 2} \end{aligned}$

Reference: L'Hôpital's rule