mathop {lim }limits_{x to 0} frac{{cos ax - c | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) Using the trigonometric identity $\cos C - \cos D = 2 \sin \left( \frac{C+D}{2} \right) \sin \left( \frac{D-C}{2} \right)$,we have:$\mathop {\lim

Vedclass · Accepted Answer

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

Vedclass · Answer

(B) Using the trigonometric identity $\cos C - \cos D = 2 \sin \left( \frac{C+D}{2} ight) \sin \left( \frac{D-C}{2} ight)$,we have:$\mathop {\lim }\limits_{x 	o 0} \frac{{\cos ax - \cos bx}}{{{x^2}}} = \mathop {\lim }\limits_{x 	o 0} \frac{{2 \sin \left( \frac{ax+bx}{2} ight) \sin \left( \frac{bx-ax}{2} ight)}}{{{x^2}}}$$= 2 \mathop {\lim }\limits_{x 	o 0} \left( \frac{\sin \left( \frac{a+b}{2}x ight)}{x} ight) \left( \frac{\sin \left( \frac{b-a}{2}x ight)}{x} ight)$$= 2 \cdot \left( \frac{a+b}{2} ight) \cdot \left( \frac{b-a}{2} ight) = 2 \cdot \frac{b^2 - a^2}{4} = \frac{b^2 - a^2}{2}$Alternatively,applying $L$-Hospital's rule:$\mathop {\lim }\limits_{x 	o 0} \frac{{\cos ax - \cos bx}}{{{x^2}}} = \mathop {\lim }\limits_{x 	o 0} \frac{{-a \sin ax + b \sin bx}}{{2x}}$$= \mathop {\lim }\limits_{x 	o 0} \frac{{-a^2 \cos ax + b^2 \cos bx}}{2} = \frac{-a^2(1) + b^2(1)}{2} = \frac{b^2 - a^2}{2}$

$\mathop {\lim }\limits_{x \to 0} \frac{{\cos ax - \cos bx}}{{{x^2}}} = $

Similar Questions

Let $f(x) = \int_0^x (t + \sin(1 - e^t)) dt, x \in R$. Then $\lim_{x \rightarrow 0} \frac{f(x)}{x^3}$ is equal to

Evaluate the limit: $\lim _{x \rightarrow 0} \frac{e^x-e^{\sin x}}{2(x-\sin x)}$

The value of $\mathop {\lim }\limits_{x \to 0} \frac{2}{x}\log (1 + x)$ is equal to

$\lim _{x \rightarrow 0} \frac{2 \sin x-\sin 2 x}{x^3}$ is equal to

The value of the limit $\mathop {\lim }\limits_{x \to 0} {\left\{ {{1^{\frac{1}{{{{\sin }^2}x}}}} + {2^{\frac{1}{{{{\sin }^2}x}}}} + \dots + {n^{\frac{1}{{{{\sin }^2}x}}}}} \right\}^{{{\sin }^2}x}}$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module