# Lim X → 0 ( Cos X + a Sin B X ) 1 / X - Mathematics

$\lim_{x \to 0} \left( \cos x + a \sin bx \right)^{1/x}$

#### Solution

$\lim_{x \to 0} \left( \cos x + a \sin bx \right)^\frac{1}{x}$
$= \lim_{x \to 0} \left[ 1 + \cos x + a \sin bx - 1 \right]^\frac{1}{x}$
$\text{ Using the theoremgiven below }:$
$If \lim_{x \to a} f\left( x \right) = \lim_{x \to a} g\left( x \right) = 0 \text{ such that } \lim_{x \to a} \frac{f\left( x \right)}{g\left( x \right)} \text{ exists, then } \lim_{x \to a} \left[ 1 + f\left( x \right) \right]^\frac{1}{g\left( x \right)} = e^\lim_{x \to a} \frac{f\left( x \right)}{g\left( x \right)} .$
$Here:$
$f\left( x \right) = \cos x + a \sin bx - 1$
$g\left( x \right) = x$
$\Rightarrow e^\lim_{x \to 0} \left[ \frac{\cos x + a \sin bx - 1}{x} \right]$
$= e^\lim_{x \to 0} \left[ \frac{b \times a \sin bx}{bx} - \frac{\left( 1 - \cos x \right)}{x} \right]$
$= e^\lim_{x \to 0} \left( \frac{ab \sin bx}{bx} - \frac{2 \sin^2 \frac{x}{2}}{x} \right)$
$= e^{ab}$

Is there an error in this question or solution?

#### APPEARS IN

RD Sharma Class 11 Mathematics Textbook
Chapter 29 Limits
Exercise 29.11 | Q 5 | Page 77

Share