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

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

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• it is a stereo-type high school limit. In all these cases you should convert $(1-\cos{k x})$ into $2 \sin^2 \frac{kx}{2}$ then simplify it with the denominator. – Arashium Oct 10 '15 at 4:40
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 $\lim_{x \to 0}\frac{\sin x}{x} = 1$.