What is the limit of $(1-cos(5x))/(sin^2(4x))$ as x approaches 0?

Question

Step by step solution would be very appreciated, I know I can use l'hopital rule but can someone explain it to me? Thanks!

Answer 1

Since $1-\cos(2u) =2 \sin^2(u) $, $1-\cos(5x) =2\sin^2(5x/2) $.

Therefore $\frac{1-\cos(5x)}{\sin^2(4x)} =\frac{2\sin^2(5x/2)}{\sin^2(4x)} $.

Now apply $\lim_{x \to 0}\frac{\sin x}{x} = 1 $.