First slide

Introduction to limits

Question

$lim_{x \to \frac{π}{2}} \frac{(1 - \tan \frac{x}{2}) (1 - \sin x)}{(1 + \tan \frac{x}{2}) (π - 2 x)^{3}}$ is equal to

Moderate

A

$\frac{1}{8}$
B

0
C

$\frac{1}{32}$
D

$\infty$

Solution

put, $x = \frac{π}{2} - h as x \to \frac{π}{2}, h \to 0$
Given,
$\begin{matrix} limit = lim_{h \to 0} \frac{1 - \tan (\frac{π}{4} + \frac{h}{2}) (1 - \cos h)}{1 + \tan (\frac{π}{2} + \frac{h}{2}) (2 h)^{3}} \\ = lim_{h \to 0} \tan (\frac{h}{2}) lim_{h \to 0} \frac{2 \cdot \sin^{2} (\frac{h}{2})}{8 h^{3}} \\ = lim_{h \to 0} \frac{1}{4} (\frac{\tan \frac{h}{2}}{2 \times \frac{h}{2}}) \times lim_{h \to 0} {(\frac{\sin \frac{h}{2}}{\frac{h}{2}})}^{2} \times \frac{1}{4} = \frac{1}{32} \end{matrix}$

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