mathop {lim }limits_{x to 0} left[ {frac{{sin | vedclass

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

Question

(C) योग-से-गुणनफल सूत्र $\sin(A+B) + \sin(A-B) = 2\sin A \cos B$ का उपयोग करने पर:$\sin(a+x) + \sin(a-x) = 2\sin a \cos x$.सीमा में इस मान को रखने पर:

Vedclass · Accepted Answer

$-\sin a$

Vedclass · Answer

(C) योग-से-गुणनफल सूत्र $\sin(A+B) + \sin(A-B) = 2\sin A \cos B$ का उपयोग करने पर:$\sin(a+x) + \sin(a-x) = 2\sin a \cos x$.सीमा में इस मान को रखने पर:$\mathop {\lim }\limits_{x 	o 0} \frac{2\sin a \cos x - 2\sin a}{x \sin x} = \mathop {\lim }\limits_{x 	o 0} \frac{2\sin a (\cos x - 1)}{x \sin x}$.$= -2\sin a \mathop {\lim }\limits_{x 	o 0} \left( \frac{1 - \cos x}{x^2} \cdot \frac{x}{\sin x} ight)$.चूंकि $\mathop {\lim }\limits_{x 	o 0} \frac{1 - \cos x}{x^2} = \frac{1}{2}$ और $\mathop {\lim }\limits_{x 	o 0} \frac{x}{\sin x} = 1$,इसलिए:$= -2\sin a \cdot \frac{1}{2} \cdot 1 = -\sin a$.

$\mathop {\lim }\limits_{x \to 0} \left[ {\frac{{\sin (x + a) + \sin (a - x) - 2\sin a}}{{x\sin x}}} \right] = $

Similar Questions

यदि $x = \log_e \left( \cot \left( \frac{\pi}{4} + \theta \right) \right)$ है,तो $\lim_{\theta \rightarrow 0} \frac{\theta}{(\sinh x)(\cosh x)} = $

$\mathop {\lim }\limits_{x \to 0} \frac{{\sin 3x + \sin x}}{x} = $

$\mathop {\lim }\limits_{\theta \to 0} \frac{{\sin 3\theta - \sin \theta }}{{\sin \theta }} = $

यदि $\mathop {\lim }\limits_{x \to 0} kx\,\text{cosec}\,x = \mathop {\lim }\limits_{x \to 0} x\,\text{cosec}\,kx$ है,तो $k = $

$\mathop {\lim }\limits_{x \to 0} \frac{x}{{\tan x}}$ का मान क्या है?

Vedclass Test Series

Exam Paper Generator

Online Exam Module