First slide

Introduction to limits

Question

$lim_{x \to 0} \frac{\tan x - x}{x^{2} \tan x}$ equals

Moderate

A

1
B

$1 / 2$
C

$1 / 3$
D

none of these

Solution

We have,
${Lim}_{=\to 0} \frac{\tan x - x}{x^{2} \tan x}$ $[\frac{0}{0} form]$
$= lim_{z \to 0} \frac{\sec^{2} x - 1}{2 x \tan x + x^{2} \sec^{2} x}$ [By L' Hospital's Rule]
$= lim_{x \to 0} \frac{2 \sec^{2} x \tan x}{2 \tan x + 4 x \sec^{2} x + 2 x^{2} \sec^{2} x \tan x}$
[By L-Hospital's Rule]
$\begin{array}{l} = lim_{x \to 0} \frac{2 \sec^{4} x + 4 \sec^{2} x \tan^{2} x}{6 \sec^{2} x + 12 x \sec^{2} x \tan x + 2 x^{2} \sec^{4} x + 4 x^{2} \sec^{2} x \tan^{2} x} \\ = \frac{2}{6} = \frac{1}{3} \end{array}$

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