The value of lim _{theta rightarrow 0} frac{ | vedclass

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

Question

(A) Given that,$ \lim _{	heta ightarrow 0} \frac{1-\cos 4 	heta}{1-\cos 6 	heta} $.Using the trigonometric identity $ 1 - \cos 2x = 2 \sin^2 x $,

Vedclass · Accepted Answer

$ \frac{4}{9} $

Vedclass · Answer

(A) Given that,$ \lim _{	heta ightarrow 0} \frac{1-\cos 4 	heta}{1-\cos 6 	heta} $.Using the trigonometric identity $ 1 - \cos 2x = 2 \sin^2 x $,we can rewrite the expression as:$ \lim _{	heta ightarrow 0} \frac{2 \sin^2(2 	heta)}{2 \sin^2(3 	heta)} = \lim _{	heta ightarrow 0} \frac{\sin^2(2 	heta)}{\sin^2(3 	heta)} $.Multiplying and dividing by $ (2 	heta)^2 $ and $ (3 	heta)^2 $ respectively:$ \lim _{	heta ightarrow 0} \left( \frac{\sin(2 	heta)}{2 	heta} ight)^2 	imes \left( \frac{3 	heta}{\sin(3 	heta)} ight)^2 	imes \frac{(2 	heta)^2}{(3 	heta)^2} $.Since $ \lim _{x ightarrow 0} \frac{\sin x}{x} = 1 $,the expression simplifies to:$ 1^2 	imes 1^2 	imes \frac{4 	heta^2}{9 	heta^2} = \frac{4}{9} $.

The value of $ \lim _{\theta \rightarrow 0} \frac{1-\cos 4 \theta}{1-\cos 6 \theta} $ is

Similar Questions

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

$\mathop {\lim }\limits_{x \to 0} \frac{{{a^{\sin x}} - 1}}{{{b^{\sin x}} - 1}} = $

$\mathop {\lim }\limits_{x \to 0} \frac{{{x^3}\cot x}}{{1 - \cos x}} = $

$\lim _{x \rightarrow 0} \frac{1-\cos x}{x^2} = $

The value of $\lim _{x \rightarrow 0} \frac{(1-\cos 2 x)(3+\cos x)}{x \tan 4 x}$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module