Question

${lim}_{x\to -1}\frac{\sqrt{\pi }-\sqrt{{\cos }^{-1}x}}{\sqrt{x+1}}$ is equal to

• A. $\frac{1}{\sqrt{2\pi }}$
• B. $\frac{1}{\sqrt{\pi }}$
• C. $\frac{1}{\sqrt{2}}$
• D. none of these

Answer 1

The correct option is D none of these

$\begin{array}{c}{lt}_{x\to -1}\frac{\sqrt{\pi }-\sqrt{{\cos }^{-1}x}}{\sqrt{x+1}}\\ {lt}_{x\to -1}\frac{\pi -{\cos }^{-1}x}{\sqrt{\left(x+1\right)}\left(\sqrt{\pi }+\sqrt{{\cos }^{-1}x}\right)}\\ \frac{1}{2\sqrt{\pi }}{lt}_{x\to -1}\frac{\pi -{\cos }^{-1}x}{\sqrt{\left(x+1\right)}}\\ Let{\cos }^{-1}x=t\\ \frac{1}{2\sqrt{\pi }}{lt}_{x\to -1}\frac{\pi -1}{\sqrt{1+\cos t}}=\frac{1}{2\sqrt{\pi }}{lt}_{t\to \pi }\frac{\pi -t}{\sqrt{2{\cos }^2\frac{t}{2}}}\\ =\frac{1}{2\sqrt{\pi }}{lt}_{t\to \pi }\frac{\pi -t}{\sqrt{2}\left|\cos \frac{t}{2}\right|}\\ =\frac{1}{2\sqrt{2}\sqrt{\pi }}{lt}_{t\to \pi }\frac{\pi -t}{\left|\cos \frac{t}{2}\right|}\\ {lt}_{x\to \pi }\frac{\pi -t}{\left|\cos \frac{t}{2}\right|}.{lt}_{t\to {\pi }^{+}}\frac{\left(\pi -t\right)}{\left|\cos \frac{t}{2}\right|}\\ \Rightarrow Limitdoesnotexist.\\ Hence,theoptionDiscorrectanswer.\end{array}$