Question

$\lim_{x \rightarrow 0}\frac{\int_{0}^{x^{2}}{(\tan^{- 1}t)^{2}dt}}{\int_{0}^{x^{4}}{\sin\sqrt{t}dt}}$is equal to -

• A. 1
• B. –1
• C. –1/2
• D. ½

Answer 1

Answer: (D)

½

Answer 2

$\lim_{x \rightarrow 0}\frac{\int_{0}^{x^{2}}{(\tan^{- 1}t)^{2}dt}}{\int_{0}^{x^{4}}{\sin\sqrt{t}dt}}$ $\left( \frac{0}{0}form \right)$

=$\lim _ { x \rightarrow 0 }$ $\frac{\lbrack(\tan^{- 1}t)^{2}\rbrack_{t = x^{2}}.\frac{d}{dx}(x^{2})}{\lbrack\sin\sqrt{t}\rbrack_{t = x^{4}}.\frac{d}{dx}(x^{4})}$

[Using L¢ Hospital's Rule]

= $\lim _ { x \rightarrow 0 }$ $\frac{(\tan^{- 1}x^{2})^{2}.2x}{\sin x^{2}.4x^{3}}$

= $\frac{1}{2}$ $\lim _ { x \rightarrow 0 }$ $\frac{\left( \frac{\tan^{- 1}x^{2}}{x^{2}} \right)^{2}}{\left( \frac{\sin x^{2}}{x^{2}} \right)} = \frac{1}{2}$.

Hence (4) is the correct answer.