Evaluate: $\lim_{x \to \pi/4}\frac{8-\sqrt{2}(\sin x+\cos x)^5}{1-\sin 2x}$ without L'Hopitals I have to find the limit of $\lim\limits_{x \to \pi/4}\frac{8-\sqrt{2}(\sin x+\cos x)^5}{1-\sin 2x}$
here is my try $\lim\limits_{x \to \pi/4}\frac{8-\sqrt{2}(\sin x+\cos x)^5}{1-\sin 2x}=\lim\limits_{x \to \pi/4}\frac{4(1-\cos^5(\frac{\pi}{4}- x))}{\sin^2 (\frac{\pi}{4}- x)}$
now observe that for $\frac{\pi}{6} \leq x \leq \frac{\pi}{3}$,
$2.5-(1-\cos^5(\frac{\pi}{4}- x))\leq\frac{1-\cos^5(\frac{\pi}{4}- x)}{\sin^2 (\frac{\pi}{4}- x)}\leq 2.5-(\sin^2 (\frac{\pi}{4}- x))$ 
then 
$\lim\limits_{x \to \pi/4}(2.5-(1-\cos^5(\frac{\pi}{4}- x)))\leq \lim\limits_{x \to \pi/4}\frac{1-\cos^5(\frac{\pi}{4}- x)}{\sin^2 (\frac{\pi}{4}- x)}\leq \lim\limits_{x \to \pi/4} (2.5-(\sin^2 (\frac{\pi}{4}- x)))$
$2.5\leq \lim\limits_{x \to \pi/4}\frac{1-\cos^5(\frac{\pi}{4}- x)}{\sin^2 (\frac{\pi}{4}- x)}\leq 2.5$ 
this implies $4\lim\limits_{x \to \pi/4}\frac{(1-\cos^5(\frac{\pi}{4}- x))}{\sin^2 (\frac{\pi}{4}- x)}=4\cdot 2.5=10$
is my answer correct? wolfram gave the same answer. is there any other simpler method? 

Observe that
$$
\cos x+\sin x=\sqrt{2}\cos(x-\pi/4)
$$
so with the substitution $x=t+\pi/4$ and noting that $\sin2x=\sin(2t+\pi/2)=\cos2t$, the limit becomes
$$
\lim_{t\to0}\frac{8-8\cos^5 t}{1-\cos2t}=
8\lim_{t\to0}\frac{1-\cos t}{1-\cos 2t}(1+\cos t+\cos^2t+\cos^3t+\cos^4t)
$$
that's easy to manage.
