Question

$\underset{x\to 0}{\lim }\frac{{\int }_0^{{X^2}^{}}\left(\sin \sqrt{t}\right)dt}{x^3}$ is equal to

• A. $\frac{2}{3}$
• B. $0$
• C. $\frac{1}{15}$
• D. $\frac{3}{2}$

Answer 1

The correct option is A

$\frac{2}{3}$

Explanations for the correct option:

Simplify the given expression.

$\begin{array}{rcl}\underset{x\to 0}{\lim }\frac{{\int }_0^{X^2}\left(\sin \sqrt{t}\right)dt}{x^3}& =& \underset{x\to 0}{\lim }\frac{{\int }_0^{X^2}\left(\sin t^{{\displaystyle \frac{1}{2}}}\right)dt}{x^3}\\ & =& \underset{x\to 0}{\lim }\frac{\left(\sin \left|x\right|\right)·2x}{3x^2}\\ & =& \underset{x\to 0}{\lim }\frac{\left(\sin \left|x\right|\right)·2}{3x}\\ & =& \frac{2}{3}·\underset{x\to 0}{\lim }\frac{\left(\sin \left|x\right|\right)}{x}\\ & =& \frac{2}{3}\times 1\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{[}\mathbf{\because }\underset{\mathbf{x}\mathbf{\to }\mathbf{0}}{\mathbf{l}\mathbf{i}\mathbf{m}}\frac{\left(sin\left|x\right|\right)}{\mathbf{x}}\mathbf{=}\mathbf{1}\mathbf{]}\\ & =& \frac{2}{3}\end{array}$

Therefore, the correct option is (A).