mathop {lim }limits_{x to 1} frac{{{x^2} - 1} | vedclass

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

Question

(C) ધારો કે પદાવલિ $L = \mathop {\lim }\limits_{x 	o 1} \frac{{{x^2} - 1}}{{{{\sin }^2}x + \cos x \cos (x + 2) - {{\cos }^2}(x + 1)}}$ છે.નિત્યસમ $\c

Vedclass · Accepted Answer

$\frac{2}{\sin 2}$

Vedclass · Answer

(C) ધારો કે પદાવલિ $L = \mathop {\lim }\limits_{x 	o 1} \frac{{{x^2} - 1}}{{{{\sin }^2}x + \cos x \cos (x + 2) - {{\cos }^2}(x + 1)}}$ છે.નિત્યસમ $\cos A \cos B = \frac{1}{2} [\cos(A+B) + \cos(A-B)]$ નો ઉપયોગ કરતા,$\cos x \cos (x+2) = \frac{1}{2} [\cos(2x+2) + \cos 2]$ મળે.તેથી,છેદ $\sin(x+1) \sin(x-1)$ માં પરિણમે છે.આમ,$L = \mathop {\lim }\limits_{x 	o 1} \frac{(x-1)(x+1)}{\sin(x+1) \sin(x-1)} = \frac{2}{\sin 2}$.

$\mathop {\lim }\limits_{x \to 1} \frac{{{x^2} - 1}}{{{{\sin }^2}x + \cos x \cos (x + 2) - {{\cos }^2}(x + 1)}}$ ની કિંમત શોધો.

Similar Questions

$\lim _{x \rightarrow 0^{-}} \frac{\sqrt{\frac{1}{2}(1-\cos ^2 x)}}{x}$ ની કિંમત શોધો.

$\mathop {\lim }\limits_{x \to 0} \frac{{5\sin x + x\cos x}}{{2\tan x - {x^2}}}$ ની કિંમત શોધો.

$\lim _{x \rightarrow 0} \frac{(1-\cos 2 x)}{x \tan 2 x+\frac{2 x}{3} \tan 3 x} = $

$\lim _{x \rightarrow 0} \frac{\cos (m x)-\cos (n x)}{x^2} =$

$\mathop {\lim }\limits_{x \to 0} \frac{1 - \cos 6x}{x} = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module