mathop {{rm{lim}}}limits_{x to 0} frac{{left( | vedclass

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

Question

(D) We need to evaluate the limit: $\mathop {\lim }\limits_{x 	o 0} \frac{{\left( {1 - \cos 2x} ight)\left( {3 + \cos x} ight)}}{{x	an 4x}}$Usin

Vedclass · Accepted Answer

$2$

Vedclass · Answer

(D) We need to evaluate the limit: $\mathop {\lim }\limits_{x 	o 0} \frac{{\left( {1 - \cos 2x} ight)\left( {3 + \cos x} ight)}}{{x	an 4x}}$Using the identity $1 - \cos 2x = 2\sin^2 x$,the expression becomes:$\mathop {\lim }\limits_{x 	o 0} \frac{{2\sin^2 x(3 + \cos x)}}{{x 	an 4x}}$Multiply and divide by $x$ and $4x$ to use standard limits $\mathop {\lim }\limits_{	heta 	o 0} \frac{\sin 	heta}{	heta} = 1$ and $\mathop {\lim }\limits_{	heta 	o 0} \frac{	an 	heta}{	heta} = 1$:$= \mathop {\lim }\limits_{x 	o 0} \left( 2 \cdot \left( \frac{\sin x}{x} ight)^2 \cdot \frac{x^2}{x \cdot \frac{	an 4x}{4x} \cdot 4x} \cdot (3 + \cos x) ight)$$= \mathop {\lim }\limits_{x 	o 0} \left( 2 \cdot \left( \frac{\sin x}{x} ight)^2 \cdot \frac{1}{4 \cdot \frac{	an 4x}{4x}} \cdot (3 + \cos x) ight)$Substituting $x = 0$:$= 2 \cdot (1)^2 \cdot \frac{1}{4 \cdot 1} \cdot (3 + \cos 0)$$= 2 \cdot \frac{1}{4} \cdot (3 + 1) = 2 \cdot \frac{1}{4} \cdot 4 = 2$

$\mathop {{\rm{lim}}}\limits_{x \to 0} \frac{{\left( {1 - \cos 2x} \right)\left( {3 + \cos x} \right)}}{{x\tan 4x}} = $

Similar Questions

$\mathop {Lim}\limits_{n \to \infty } \frac{{{1^2}n + {2^2}(n - 1) + {3^2}(n - 2) + \dots + {n^2} \cdot 1}}{{{1^3} + {2^3} + {3^3} + \dots + {n^3}}}$ is equal to :

The value of $\mathop {\lim }\limits_{x \to \infty } \frac{{{x^2}\sin \frac{1}{x} - x}}{{1 - |x|}}$ is

The value of $\mathop {\text{Limit}}\limits_{x \to \infty } \frac{{\cot ^{ - 1}}\left( {{x^{ - a}}\log _a x} \right)}{{\sec ^{ - 1}}\left( {{a^x}\log _x a} \right)}$ for $a > 1$ is equal to

Find the limit: $\mathop {\lim }\limits_{x \to 3} [x(x+1)]$

$\mathop {\lim }\limits_{n \to \infty } \frac{{\sqrt n }}{{\sqrt n + \sqrt {n + 1} }} = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module