lim _{x rightarrow 0} frac{x tan 2 x-2 x tan | vedclass

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

Question

(C) આપણી પાસે પદાવલિ $L = \lim _{x ightarrow 0} \frac{x 	an 2 x-2 x 	an x}{(1-\cos 3 x)(\operatorname{cosec} x-\cot x)^2}$ છે.પ્રથમ,છેદના પદ $(\op

Vedclass · Accepted Answer

$\frac{16}{9}$

Vedclass · Answer

(C) આપણી પાસે પદાવલિ $L = \lim _{x ightarrow 0} \frac{x 	an 2 x-2 x 	an x}{(1-\cos 3 x)(\operatorname{cosec} x-\cot x)^2}$ છે.પ્રથમ,છેદના પદ $(\operatorname{cosec} x-\cot x) = \frac{1-\cos x}{\sin x} = 	an(x/2)$ ને સરળ બનાવો.તેથી,$(\operatorname{cosec} x-\cot x)^2 = 	an^2(x/2) \approx \frac{x^2}{4}$ જ્યારે $x ightarrow 0$.વળી,$1-\cos 3x = 2 \sin^2(\frac{3x}{2}) \approx \frac{9x^2}{2}$.છેદ $\frac{9x^2}{2} \cdot \frac{x^2}{4} = \frac{9x^4}{8}$ બને છે.હવે,અંશને સરળ બનાવો: $x 	an 2x - 2x 	an x = 2x^4$.આમ,$L = \lim _{x ightarrow 0} \frac{2x^4}{9x^4/8} = \frac{16}{9}$.

$\lim _{x \rightarrow 0} \frac{x \tan 2 x-2 x \tan x}{(1-\cos 3 x)(\operatorname{cosec} x-\cot x)^2}=$

Similar Questions

જો $f(x) = -(\sin^2 x + \cos^5 x)$ હોય,તો $\lim_{x \rightarrow 0} \frac{f'(x)}{x}$ શોધો.

$\mathop {\lim }\limits_{\theta \to 0} \frac{{5\theta \cos \theta - 2\sin \theta }}{{3\theta + \tan \theta }} = $

$\lim _{x}$ ${\rightarrow 0} \frac{\tan \left(\left[-\pi^2\right] x^2\right)-x^2 \tan \left(\left[-\pi^2\right]\right)}{\sin ^2 x}$ ની કિંમત શોધો.

$\mathop {\lim }\limits_{\theta \to 0} \frac{{1 - \cos \theta }}{{{\theta ^2}}} = $

$\lim _{x}$ ${\rightarrow 0} \left( \left( \frac{1-\cos ^2(3 x)}{\cos ^3(4 x)} \right) \left( \frac{\sin ^3(4 x)}{(\log _e(2 x+1))^5} \right) \right)$ ની કિંમત $.........$ છે.

Vedclass Test Series

Exam Paper Generator

Online Exam Module