conceptual doubt in finding $\lim \limits_{x \to \tfrac{\pi}{4}} \frac{4\sqrt 2 -\left ( \cos x + \sin x \right )^5}{1-\sin {2x}}$ While calculating the aforementioned limits I did the following.
$$\lim \limits_{x \to \tfrac{\pi}{4}} \frac{4\sqrt 2 -\left ( \cos x + \sin x \right )^5}{1-\sin {2x}}\\ =
\lim \limits_{x \to \tfrac{\pi}{4}} \frac{4\sqrt 2 -4\sqrt2 \left (\frac{1}{\sqrt 2} \cos x +\frac{1}{\sqrt 2} \sin x \right )^5}{1-\sin {2x}}\\=
\lim \limits_{x \to \tfrac{\pi}{4}} \frac{4\sqrt {2} \left ( 1 - \sin^5 (x + \frac{\pi}{4}) \right )}{1-\sin {2x}}\\
=\lim \limits_{x \to \tfrac{\pi}{4}} \frac{4\sqrt {2} \left ( 1 - \sin (x + \frac{\pi}{4}) \right ) \left(1 +\sin (x + \frac{\pi}{4}) +\sin^2 (x + \frac{\pi}{4}) +\sin^3 (x + \frac{\pi}{4}) +\sin^4 (x + \frac{\pi}{4}) \right)}{1-\sin {2x}}\\$$
As $x$ approaches $\frac{\pi}{4}$, the functions $\sin (x + \frac{\pi}{4})$ and $\sin 2x$ become close enough. So, I canceled them($1-\sin {2x}$ and $ 1 - \sin (x + \frac{\pi}{4})$ ) and the limit turns out to be $20\sqrt2$. 
I made the above claim after comparing the graphs. But the fact that bothers me is L'Hospital's rule yielded the answer $5\sqrt2$ (and it is the correct answer to the problem)
Can anyone please explain?

 A: Make life easier using $x=y+\frac{\pi}{4}$
$$\lim \limits_{x \to \tfrac{\pi}{4}} \frac{4\sqrt 2 -\left ( \cos (x) + \sin (x) \right )^5}{1-\sin ({2x})}$$ Combine
$$\cos (x) + \sin (x)=\sqrt{2} \cos (y)$$ So, in terms of $y$, the expression reduces to
$$\frac{4\sqrt 2 -4 \sqrt{2} \cos ^5(y) )}{2 \sin ^2(y)}=2\sqrt 2\,\frac{1-  \cos ^5(y) }{ \sin ^2(y)}$$
Now Taylor series
$$\cos(y)=1-\frac{y^2}{2}+\frac{y^4}{24}+O\left(y^6\right)$$
$$\cos^5(y)=1-\frac{5 y^2}{2}+\frac{65 y^4}{24}+O\left(y^6\right)$$
Just finish
A: 
As $x$ approaches $\frac{\pi}{4}$ the functions $\sin (x + \frac{\pi}{4})$ and $\sin 2x$ become close enough

When we say "close enough", we are saying that $\lim_{x\to \pi/4} \dfrac{\sin (x + \frac{\pi}{4})}{\sin 2x}=1$. However, the limit
$\lim_{x\to \pi/4} \dfrac{1-\sin (x + \frac{\pi}{4})}{1-\sin 2x}=\dfrac{1}{4}\neq 1$
So we cannot cancel them simply.
We can view it from the perspective of Taylor Series:
$$
\frac{\sin \left(x+\frac{\pi }{4}\right)}{\sin (2 x)}=1+\frac{3}{2} \left(x-\frac{\pi }{4}\right)^2+\frac{19}{8} \left(x-\frac{\pi }{4}\right)^4+O\left(\left(x-\frac{\pi }{4}\right)^5\right) \\
\frac{1-\sin \left(x+\frac{\pi }{4}\right)}{1-\sin (2 x)}=\frac{1}{4}+\frac{1}{16} \left(x-\frac{\pi }{4}\right)^2+\frac{1}{96} \left(x-\frac{\pi }{4}\right)^4+O\left(\left(x-\frac{\pi }{4}\right)^5\right)
$$
