lim _{x} {rightarrow 0} frac{x+2 sin x+3 tan | vedclass

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

Question

(A) Let $f(x) = x+2 \sin x+3 	an x-	an ^3 x$ and $g(x) = \sqrt{x^2+2 \sin x+	an x+3}-\sqrt{\sin ^2 x-2 	an x-x+3}$.As $x ightarrow 0$,$f(0) = 0+

Vedclass · Accepted Answer

$2 \sqrt{3}$

Vedclass · Answer

(A) Let $f(x) = x+2 \sin x+3 	an x-	an ^3 x$ and $g(x) = \sqrt{x^2+2 \sin x+	an x+3}-\sqrt{\sin ^2 x-2 	an x-x+3}$.As $x ightarrow 0$,$f(0) = 0+0+0-0 = 0$ and $g(0) = \sqrt{3}-\sqrt{3} = 0$. This is a $\frac{0}{0}$ form.We apply $L$'Hopital's rule by differentiating the numerator and denominator.Numerator derivative: $f'(x) = 1+2 \cos x+3 \sec ^2 x-3 	an ^2 x \sec ^2 x$.At $x=0$,$f'(0) = 1+2(1)+3(1)-0 = 6$.Denominator derivative: $g'(x) = \frac{2x+2 \cos x+\sec ^2 x}{2 \sqrt{x^2+2 \sin x+	an x+3}} - \frac{2 \sin x \cos x-2 \sec ^2 x-1}{2 \sqrt{\sin ^2 x-2 	an x-x+3}}$.At $x=0$,$g'(0) = \frac{0+2+1}{2 \sqrt{3}} - \frac{0-2-1}{2 \sqrt{3}} = \frac{3}{2 \sqrt{3}} - \frac{-3}{2 \sqrt{3}} = \frac{6}{2 \sqrt{3}} = \frac{3}{\sqrt{3}} = \sqrt{3}$.The limit is $\frac{f'(0)}{g'(0)} = \frac{6}{\sqrt{3}} = 2 \sqrt{3}$.

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

Similar Questions

$\lim _{x \rightarrow 3^{-}} \frac{x^3-3 x^2-4 x+12}{2 x^3-7 x^2+2 x+3} = $

$\lim _{x \rightarrow 1} \frac{\log x}{1-x} = $

$\mathop {\lim }\limits_{x \to 1} \frac{{\log x}}{{x - 1}} = $

The value of the limit of $\frac{x^3 - x^2 - 18}{x - 3}$ as $x$ tends to $3$ is

If $\mathop {\lim }\limits_{x \to a} \frac{{{a^x} - {x^a}}}{{{x^x} - {a^a}}} = - 1$,then

Vedclass Test Series

Exam Paper Generator

Online Exam Module