Evaluate: $\lim_{x \to \pi/4}\frac{8-\sqrt{2}(\sin x+\cos x)^5}{1-\sin 2x}$ without L'Hopitals

Question

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?

Answer 1

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.

Answer 2

$\lim\limits_{x \to \pi/4}\frac{8-\sqrt{2}(\sin x+\cos x)^5}{1-\sin 2x}$

$= \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, $\color{blue}{a-b= \dfrac{a^2 - b^2 }{a+b}}$

$= \lim_{x \to \pi/4}\dfrac{\sqrt 2. 20 ( 2-(\sin ^2 x + \cos^2 x+\sin 2x) )}{(\sin x + \cos x + \sqrt 2)(1- \sin 2x)}$

$= \lim _{x \to \pi/4}\dfrac{\sqrt 2 .20}{2\sqrt 2}$

$= 10$