<s>[INST] 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? 
 [/INST] $\lim\limits_{x \to \pi/4}\frac{8-\sqrt{2}(\sin x+\cos x)^5}{1-\sin 2x}$
$= \lim_{x \to \frac{\pi}4}\dfrac{\sqrt2(\sqrt2^5 - (\sin x + \cos x)^5)}{1- \sin 2x} $
Using, 
$\color{blue}{\lim _{x \to a}\frac{x^n -a^n}{x-a}=na^{n-1}}$
$= \lim _{x \to \pi/4}\dfrac{\sqrt 2 (5.\sqrt2^{4})(\sqrt 2 -(\sin x +\cos x))}{1- \sin 2x}$
Using, 
$\color{blue}{a-b= \dfrac{a^2 - b^2 }{a+b}}$
$= \lim_{x \to \pi/4}\dfrac{\sqrt 2. 20 ( 2-(\sin ^2 x + \cos^2 x+\sin 2x) )}{(\sin x + \cos x + \sqrt 2)(1- \sin 2x)}$
$= \lim _{x \to \pi/4}\dfrac{\sqrt 2 .20}{2\sqrt 2}$
$= 10$
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