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

$\lim _{x \rightarrow 0} \frac{(1-\cos 2 x) \cdot \sin 5 x}{x^2 \sin 3 x}$ is
  1. $\frac{10}{3}$
  2. $\frac{5}{3}$
  3. $\frac{5}{6}$
  4. $\frac{2}{3}$

Solution

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

Asked in: MHT CET 2023 (09 May Shift 2)

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