Question

$\underset{x\to 0}{\lim }\frac{(1-\cos 2x)\sin 5x}{x^2\sin 3x}=$

• A. $\frac{10}{3}$
• B. $\frac{3}{10}$
• C. $\frac{6}{5}$
• D. $\frac{5}{6}$

Answer 1

The correct option is A

$\frac{10}{3}$

Explanation for the correct option:

Finding the value of the given limit:

$$\begin{gathered} \underset{x\to 0}{\lim }\frac{(1-\cos 2x)\sin 5x}{x^2\sin 3x} \\ =\underset{x\to 0}{\lim }\frac{(1-\cos 2x)}{x^2}\times \underset{x\to 0}{\lim }\frac{\sin 5x}{5x}\times \underset{x\to 0}{\lim }5x\times \underset{x\to 0}{\lim }\frac{3x}{\sin 3x}\times \underset{x\to 0}{\lim }\frac{1}{3x}...[\because \mathrm{Splitting}\mathrm{the}\mathrm{terms}] \\ =\underset{x\to 0}{\lim }\frac{2{\sin }^{2}x}{x^2}\times \underset{x\to 0}{\lim }\frac{\sin 5x}{5x}\times \underset{x\to 0}{\lim }\frac{3x}{\sin 3x}\times \underset{x\to 0}{\lim }\frac{5x}{3x}...\left[\because \mathbf{1}\mathbf{}\mathbf{-}\mathbf{}\mathit{c}\mathit{o}\mathit{s}\mathbf{}\mathbf{2}\mathit{x}\mathbf{=}\mathbf{2}\mathit{s}\mathit{i}{\mathit{n}}^{\mathbf{2}}\mathit{x}\right] \\ =\underset{x\to 0}{\lim }\frac{5\times 2}{3}{\left(\frac{\sin \left(x\right)}{x}\right)}^2\left(\frac{\sin 5x}{5x}\right)\left(\frac{3x}{\sin 3x}\right) \end{gathered}$$

Applying the limits,

$$\begin{gathered} =\frac{10}{3}{\left(1\right)}^2\left(1\right)\left(1\right)\left[\because \underset{\theta \to 0}{\lim }\frac{\sin \left(\theta \right)}{\theta }=1\right] \\ =\frac{10}{3} \end{gathered}$$

Therefore, the correct answer is option (A).