$$\lim_{x\to\tfrac{\pi}{4}} \frac{(\cos x + \sin x)^3 - 2\sqrt2}{1 - \sin 2x}$$
I tried solving this question without using L'Hospital's Rule but couldn't find a satisfactory way of approach to solve this particular question. I would be highly obliged if I could get a solution from anyone of you regarding how we can evaluate this limit without using L'Hospital's Rule. Thank you in advance!
Hint: Substitute $y=\frac\pi4-x$, so that $\cos(x)+\sin(x)=\sqrt2\cos(y)$ and $\sin(2x)=\cos(2y)$. These follow from the identities, $$\cos(a+b)=\cos a\cos b-\sin a\sin b\\ \sin(a+b)=\sin a\cos b+\cos a\sin b$$
In particular,
$$\cos x+\sin x = \sqrt2 \left(\cos\frac\pi4\cos x + \sin\frac\pi4 \sin x\right) = \sqrt 2 \cos\left(\frac\pi4-x\right) = \sqrt2 \cos y \\ \sin(2x) = \sin\left(2\left(\frac\pi4-y\right)\right) = \sin\left(\frac\pi2 - 2y\right) = \sin\frac\pi2 \cos(2y) - \cos\frac\pi2 \sin(2y) = \cos(2y)$$
Now the limit is
$$\begin{align*} L &= \lim_{x\to\tfrac\pi4} \frac{(\cos x+\sin x)^3 - 2\sqrt2}{1 - \sin(2x)} \\ &= 2\sqrt2 \lim_{y\to0} \frac{\cos^3y-1}{1 - \cos(2y)} \end{align*}$$