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This question already has an answer here:

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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marked as duplicate by Eclipse Sun, lab bhattacharjee limits Oct 10 '15 at 4:45

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  • $\begingroup$ You must show your efforts so that someone can help you $\endgroup$ – Shailesh Oct 10 '15 at 4:36
  • $\begingroup$ 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. $\endgroup$ – Arashium Oct 10 '15 at 4:40
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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 $.

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