# How to solve this without L'Hopital's Rule?

$$\lim_{x \to \frac{\pi}{4}}\frac{\cos(4x) +1}{ \sin(8x)}$$

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

• Hint: Use multiple angle formula to convert cos(4x) into cos(2x),then it would be standard limit. Jan 27 at 9:23

$$\frac{\cos(4x) +1}{ \sin(8x)}=\frac {2\cos^{2}(2x)} {2\sin (4x)\cos (4x)}=\frac {2\cos^{2}(2x)} {4\sin (2x)\cos (2x)\cos (4x)}$$ Note that one factor of $$\cos (2x)$$ cancels and $$\cos (4x) \to -1$$. Can you finish?