Lim X → 0 + { 1 + Tan 2 √ X } 1 / 2 X - Mathematics

$\lim_{x \to 0^+} \left\{ 1 + \tan^2 \sqrt{x} \right\}^{1/2x}$

Solution

$\lim_{x \to 0^+} \left[ 1 + \tan^2 \sqrt{x} \right] {}^\frac{1}{2x}$
$\text{ Using the theorem given below }:$
$\text{ If } \lim_{x \to a} f\left( x \right) = \lim_{x \to a} g\left( x \right) = 0 \text{ such that } \lim_{x \to a} \frac{f\left( x \right)}{g\left( x \right)} \text{ exists, then } \lim_{x \to a} \left[ 1 + f\left( x \right) \right]^\frac{1}{g\left( x \right)} = e^\lim_{x \to a} \frac{f\left( x \right)}{g\left( x \right)} .$
$\text{ Here }:$
$f\left( x \right) = \tan^2 \sqrt{x}$
$g\left( x \right) = 2x$
$\Rightarrow e^\lim_{x \to 0^+} \left( \frac{\tan^2 \sqrt{x}}{2x} \right)$
$= e^\lim_{x \to 0^+} \left( \frac{\tan \sqrt{x}}{\sqrt{x}} \right) \times \left( \frac{\tan \sqrt{x}}{\sqrt{x}} \right) \times \frac{1}{2}$
$= e^{1 \times 1 \times \frac{1}{2}}$
$= \sqrt{e}$

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APPEARS IN

RD Sharma Class 11 Mathematics Textbook
Chapter 29 Limits
Exercise 29.11 | Q 2 | Page 76