The value of $\lim _{x \rightarrow 0} \frac{(1-\cos 2 x) \sin 5 x}{x^2 \sin 3 x}$ is
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The value of $\lim _{x \rightarrow 0} \frac{(1-\cos 2 x) \sin 5 x}{x^2 \sin 3 x}$ is

(A) $10 / 3$

(B) $3 / 10$

(C) $6 / 5$

(D) $5 / 6$

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SOLUTION —

$\lim _{x \rightarrow 0} \frac{(1-\cos 2 x) \sin 5 x}{x^2 \sin 3 x}$

$\begin{array}{l}=\lim _{x \rightarrow 0} \frac{1-\cos 2 x}{x^2} \times \frac{\sin 5 x}{5 \cdot \frac{x}{5}} \times \frac{3 x}{3 \sin 3 x} \\=\lim _{x \rightarrow 0} \frac{2 \sin ^2 x}{x^2} \times 5 \times \frac{1}{3}=\frac{10}{3} \quad\left(\because \lim _{x \rightarrow 0} \frac{\sin x}{x}=1\right)\end{array}$

So, The correct option will be (A).

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