mathop {lim }limits_{x to 0} frac{{x + 2sin x | vedclass

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

Question

(A) Let $L = \mathop {\lim }\limits_{x 	o 0} \frac{{x + 2\sin x}}{{\sqrt {{x^2} + 2\sin x + 1} - \sqrt {{{\sin }^2}x - x + 1} }}$.Rationalizing the d

Vedclass · Accepted Answer

$2$

Vedclass · Answer

(A) Let $L = \mathop {\lim }\limits_{x 	o 0} \frac{{x + 2\sin x}}{{\sqrt {{x^2} + 2\sin x + 1} - \sqrt {{{\sin }^2}x - x + 1} }}$.Rationalizing the denominator:$L = \mathop {\lim }\limits_{x 	o 0} \frac{{(x + 2\sin x)(\sqrt {{x^2} + 2\sin x + 1} + \sqrt {{{\sin }^2}x - x + 1} )}}{({x^2} + 2\sin x + 1) - ({{\sin }^2}x - x + 1)}$$L = \mathop {\lim }\limits_{x 	o 0} \frac{{(x + 2\sin x)(\sqrt {{x^2} + 2\sin x + 1} + \sqrt {{{\sin }^2}x - x + 1} )}}{{x^2 + x + 2\sin x - {{\sin }^2}x}}$Dividing numerator and denominator by $x$:$L = \mathop {\lim }\limits_{x 	o 0} \frac{{(1 + 2\frac{\sin x}{x})(\sqrt {{x^2} + 2\sin x + 1} + \sqrt {{{\sin }^2}x - x + 1} )}}{{x + 1 + 2\frac{\sin x}{x} - \sin x \frac{\sin x}{x}}}$As $x 	o 0$,$\frac{\sin x}{x} 	o 1$ and $\sin x 	o 0$:$L = \frac{{(1 + 2(1))(\sqrt {0 + 0 + 1} + \sqrt {0 - 0 + 1} )}}{{0 + 1 + 2(1) - 0(1)}} = \frac{{3(1 + 1)}}{3} = 2$.Hence,the correct answer is option $(A)$.

$\mathop {\lim }\limits_{x \to 0} \frac{{x + 2\sin x}}{{\sqrt {{x^2} + 2\sin x + 1} - \sqrt {{{\sin }^2}x - x + 1} }}$ is

Similar Questions

Evaluate the given limit: $\mathop {\lim }\limits_{x \to 0} \frac{ax + x \cos x}{b \sin x}$

Evaluate the given limit: $\mathop {\lim }\limits_{x \to \pi } \left(x-\frac{22}{7}\right)$

$\lim _{x \rightarrow \infty}\left[\frac{8 x^2+5 x+3}{2 x^2-7 x-5}\right]^{\frac{4 x+3}{8 x-1}} = $

$\lim _{x \rightarrow 0} \frac{3^{\sin x}-2^{\tan x}}{\sin x}=$

$\lim _{x \rightarrow a} \frac{\sqrt{a+2 x}-\sqrt{3 x}}{\sqrt{3 a+x}-2 \sqrt{x}} = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module