# Evaluate the following limits: lim x → 0 cos a x − cos b x cos c x − 1 - Mathematics

Evaluate the following limits:

$\lim_{x \to 0} \frac{\cos ax - \cos bx}{\cos cx - 1}$

#### Solution

$\lim_{x \to 0} \frac{\cos ax - \cos bx}{\cos cx - 1}$

$= \lim_{x \to 0} \frac{- 2\sin\left( \frac{ax + bx}{2} \right)\sin\left( \frac{ax - bx}{2} \right)}{- 2 \sin^2 \frac{cx}{2}} \left( 1 - \cos2\theta = 2 \sin^2 \theta \right)$

$= \lim_{x \to 0} \frac{\left( \frac{a + b}{2} \right)x \times \frac{\sin\left( \frac{a + b}{2} \right)x}{\left( \frac{a + b}{2} \right)x} \times \left( \frac{a - b}{2} \right)x \times \frac{\sin\left( \frac{a - b}{2} \right)x}{\left( \frac{a - b}{2} \right)x}}{\frac{c^2 x^2}{4} \times \frac{\sin^2 \frac{cx}{2}}{\frac{c^2 x^2}{4}}}$

$= \frac{\left( a + b \right)\left( a - b \right)}{c^2} \times \frac{\lim_{x \to 0} \frac{\sin\left( \frac{a + b}{2} \right)x}{\left( \frac{a + b}{2} \right)x} \times \lim_{x \to 0} \frac{\sin\left( \frac{a - b}{2} \right)x}{\left( \frac{a - b}{2} \right)x}}{\left( \lim_{x \to 0} \frac{\sin\frac{cx}{2}}{\frac{cx}{2}} \right)^2}$

$= \frac{a^2 - b^2}{c^2} \times \frac{1 \times 1}{1} \left( \lim_\theta \to 0 \frac{\sin\theta}{\theta} = 1 \right)$

$= \frac{a^2 - b^2}{c^2}$

Is there an error in this question or solution?

#### APPEARS IN

RD Sharma Class 11 Mathematics Textbook
Chapter 29 Limits
Exercise 29.7 | Q 61 | Page 51