# Evaluating $$\displaystyle\lim_{{{x}\to{0}}}{\frac{{{1}-{\cos{{x}}}}}{{{\left({2}{x}\right)}^{{2}}}}}$$ I have this question in my

Evaluating $\underset{x\to 0}{lim}\frac{1-\mathrm{cos}x}{{\left(2x\right)}^{2}}$
I have this question in my book and the answer amounts to $\frac{1}{4}$. I don't see how, and I think it's an error. Can you confirm it is an error?
$\underset{x\to 0}{lim}\frac{1-\mathrm{cos}x}{{\left(2x\right)}^{2}}$
I assume it would need to be ${\mathrm{cos}}^{2}x$ instead.
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membatas0v2v
It's indeed not correct. Simply use l'Hopital rule allow you to conclude that
$\underset{x\to 0}{lim}\frac{1-\mathrm{cos}\left(x\right)}{{\left(2x\right)}^{2}}=\underset{x\to 0}{lim}\frac{1-\mathrm{cos}\left(x\right)}{4{x}^{2}}=\underset{x\to 0}{lim}\frac{\mathrm{sin}\left(x\right)}{8x}=\frac{1}{8}$