First slide

Introduction to limits

Question

$lim_{x \to - 1} \frac{\sqrt{π} - \sqrt{\cos^{- 1} x}}{\sqrt{x + 1}}$ is given by

Moderate

A

$\frac{1}{\sqrt{π}}$
B

$\frac{1}{\sqrt{2 π}}$
C

1
D

0

Solution

Let $\cos^{- 1} x = y$ Then, $x \to - 1^{+} \Rightarrow y \to π^{-}$
$∴ lim_{x \to - 1^{+}} \frac{\sqrt{π} - \sqrt{\cos^{- 1} x}}{\sqrt{x + 1}}$
$\begin{aligned} = lim_{y \to π^{-}} \frac{\sqrt{π} - \sqrt{y}}{\sqrt{1 + \cos y}} \\ = lim_{y \to π^{-}} \frac{\sqrt{π} - \sqrt{y}}{\sqrt{2} \cos y / 2} \end{aligned}$
$= lim_{y \to π^{-}} \frac{\sqrt{π} - \sqrt{y}}{\sqrt{2} \sin (\frac{π}{2} - \frac{y}{2})} \times \frac{(\frac{π}{2} - \frac{y}{2})}{(\frac{π}{2} - \frac{y}{2})}$
$= lim_{y \to π^{-}} \frac{1}{\frac{\sqrt{2}}{2} (\sqrt{π} + \sqrt{y})} \cdot \frac{1}{\{\frac{\sin (\frac{π}{2} - \frac{y}{2})}{\frac{π}{2} - \frac{y}{2}}\}}\} = \frac{1}{\sqrt{2 π}}$

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