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

$\lim _{x \rightarrow 2}\left(\frac{5^x+5^{3-x}-30}{5^{3-x}-5^{\frac{x}{2}}}\right)=$

$\mathop {\lim }\limits_{x \to 0} \frac{|x|}{x} = $

$\mathop {\lim }\limits_{x \to \infty } {\left( {1 - \frac{4}{{x - 1}}} \right)^{3x - 1}} = $

$\lim _{x \rightarrow 0} \frac{\sqrt{x^2+100}-10}{x^2} = $

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

Vedclass Test Series

Exam Paper Generator

Online Exam Module