Prove the following
Question:

$\lim _{x \rightarrow 0} \frac{\sin ^{2} 2 x}{\sin ^{2} 4 x}$

Solution:

Given $\lim _{x \rightarrow 0} \frac{\sin ^{2} 2 x}{\sin ^{2} 4 x}$

Multiply and divide both numerator and denominator by $4 x^{2} / 16 x^{2}$ then we get

$\Rightarrow \lim _{x \rightarrow 0} \frac{\sin ^{2} 2 x}{\sin ^{2} 4 x}=\lim _{x \rightarrow 0} \frac{\left(\sin ^{2} 2 x\right) / 4 x^{2}}{\left(\sin ^{2} 4 x\right) / 16 x^{2}} \times \frac{4 x^{2}}{16 x^{2}}$

On simplifying

$\Rightarrow \lim _{x \rightarrow 0}\left[\frac{\left(\frac{\sin 2 x}{2 x}\right)^{2}}{\left(\frac{\sin 4 x}{4 x}\right)^{2}}\right] \times \frac{4}{16}$

Now as $\lim _{x \rightarrow 0} \frac{\sin x}{x}=1$

$\lim _{x \rightarrow 0}\left[\frac{\left(\frac{\sin 2 x}{2 x}\right)^{2}}{\left(\frac{\sin 4 x}{4 x}\right)^{2}}\right] \times \frac{4}{16}=\frac{4}{16}$

$\lim _{x \rightarrow 0} \frac{\sin ^{2} 2 x}{\sin ^{2} 4 x}=\frac{4}{16}$

$=1 / 4$

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