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

$\mathop {\lim }\limits_{x \to - 2} \frac{{{{\sin }^{ - 1}}(x + 2)}}{{{x^2} + 2x}}$ is equal to

$\mathop {\lim }\limits_{x \to 0} \frac{{{{(1 + x)}^{1/2}} - {{(1 - x)}^{1/2}}}}{x} = $

Let $f(x) = x^{6} + 2x^{4} + x^{3} + 2x + 3$,$x \in R$. Then the natural number $n$ for which $\lim_{x \rightarrow 1} \frac{x^{n} f(1) - f(x)}{x - 1} = 44$ is ...... .

The value of $\mathop {\lim }\limits_{x \to 0} \frac{{{e^x} - {e^{ - x}}}}{{\sin x}}$ is

$\lim _{x \rightarrow 2} \frac{\sqrt{1+4 x}-\sqrt{3+3 x}}{x^3-8} = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module