$$\lim_{x \to \frac{\pi}{4}}\frac{\cos(4x) +1}{ \sin(8x)}$$
I tried writing $$\sin(8x)=8\sin(x)\cos(x)\cos(2x)\cos(4x)$$
and substitute x with $$x=y+\frac{\pi}{4}$$ while y approaches 0
2026-04-09 11:54:05.1775735645
How to solve this without L'Hopital's Rule?
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$$\frac{\cos(4x) +1}{ \sin(8x)}=\frac {2\cos^{2}(2x)} {2\sin (4x)\cos (4x)}=\frac {2\cos^{2}(2x)} {4\sin (2x)\cos (2x)\cos (4x)}$$ Note that one factor of $\cos (2x)$ cancels and $\cos (4x) \to -1$. Can you finish?