Question

The value of $\underset{x\to 0}{lim}\frac{\frac{d}{dx}{\int }_0^{x^2}\sqrt{\cos t}dt}{1-\sqrt{\cos x}}$ is?

• A. $0$
• B. $11$
• C. $10$
• D. $12$

Answer 1

The correct option is A $0$

$\begin{array}{c}{lim}_{x\to 0}\frac{\frac{d}{dx}{\int }_0^{x^2}\sqrt{\cos tdt}}{1-\sqrt{\cos x}}\\ \Rightarrow {lim}_{x\to 0}\frac{\sqrt{\cos x^2\times 2x-0}}{1-\sqrt{\cos x}}\\ \Rightarrow {lim}_{x\to 0}\frac{2x\sqrt{\cos x^2}}{1-\sqrt{\cos x}}\left[\because \frac{0}{0}form\right]\\ UsingHospitalrule\\ \Rightarrow {lim}_{x\to 0}\frac{\left[x\frac{1}{2\sqrt{\cos x^2}}\right]}{\frac{\sin x^2}{2\sqrt{\cos x}}}\\ \Rightarrow {lim}_{x\to 0}\frac{4\times \left[x^2\frac{\sin x^2}{\sqrt{\cos x^2}}+\sqrt{\cos x^2}\right]}{\frac{\sin x}{\sqrt{\cos x}}}\\ \Rightarrow {lim}_{x\to 0}\frac{4\times \left[x\frac{\sin x^2}{\sqrt{\cos x^2}}+\sqrt{\cos x^2}\right]}{\frac{\sin x}{x}\times \frac{1}{\sqrt{\cos x}}}\\ {lim}_{x\to 0}\frac{\left[\frac{x\sin x^2}{\sqrt{\cos x^2}}+\sqrt{\cos x^2}\right]}{\frac{1}{\sqrt{\cos x}}}\left[\because {lim}_{x\to 0}\frac{\sin x}{x}=1\right]\\ {lim}_{x\to 0}=4\left[\frac{0+1}{\frac{1}{1}}\right]=4Ans.\end{array}$