Q128 / 200 · Manipal (MET) 2012 · Mathematics
The value of $\lim _{x \rightarrow 0} \frac{\int_0^{x^2} \cos ^2 t d t}{x \sin x}$ is
- A2
- Correct answerB1
- C0
- DNone of these
Solution
$\lim _{x \rightarrow 0} \frac{\int_0^{x^2} \cos ^2 t d t}{x \sin x}$
Using L Hospital's rule, we get $\begin{aligned} & =\lim _{x \rightarrow 0} \frac{\cos ^2\left(x^2\right) \cdot 2 x-0}{x \cos x+\sin x} \\ & =\lim _{x \rightarrow 0} \frac{2 \cos ^2\left(x^2\right)}{\cos x+\frac{\sin x}{x}} \\ & =\frac{2}{2}=1 \end{aligned}$
Using L Hospital's rule, we get $\begin{aligned} & =\lim _{x \rightarrow 0} \frac{\cos ^2\left(x^2\right) \cdot 2 x-0}{x \cos x+\sin x} \\ & =\lim _{x \rightarrow 0} \frac{2 \cos ^2\left(x^2\right)}{\cos x+\frac{\sin x}{x}} \\ & =\frac{2}{2}=1 \end{aligned}$
Hence, the correct answer is option: 'B'.