lim _{x rightarrow 2} frac{sqrt{1+4 x}-sqrt{3 | vedclass

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

Question

(A) Given limit is $\lim _{x \rightarrow 2} \frac{\sqrt{1+4 x}-\sqrt{3+3 x}}{x^3-8}$.Since it is a $\frac{0}{0}$ indeterminate form,we apply $L'H\hat{

Vedclass · Accepted Answer

$\frac{1}{72}$

Vedclass · Answer

(A) Given limit is $\lim _{x \rightarrow 2} \frac{\sqrt{1+4 x}-\sqrt{3+3 x}}{x^3-8}$.Since it is a $\frac{0}{0}$ indeterminate form,we apply $L'H\hat{o}pital's$ rule by differentiating the numerator and denominator with respect to $x$:$\lim _{x \rightarrow 2} \frac{\frac{d}{dx}(\sqrt{1+4 x}-\sqrt{3+3 x})}{\frac{d}{dx}(x^3-8)} = \lim _{x \rightarrow 2} \frac{\frac{4}{2\sqrt{1+4x}} - \frac{3}{2\sqrt{3+3x}}}{3x^2}$.Substituting $x = 2$:$= \frac{\frac{4}{2\sqrt{9}} - \frac{3}{2\sqrt{9}}}{3(2^2)} = \frac{\frac{4}{6} - \frac{3}{6}}{12} = \frac{\frac{1}{6}}{12} = \frac{1}{72}$.

$\lim _{x \rightarrow 2} \frac{\sqrt{1+4 x}-\sqrt{3+3 x}}{x^3-8} = $

Similar Questions

$\lim _{x \rightarrow 0} \frac{\tan x - \sin x}{x^3}$ is equal to

If $f(a)=2, f^{\prime}(a)=1, g(a)=-1, g^{\prime}(a)=2$,then as $x$ approaches $a$,the limit of $\frac{g(x) f(a)-g(a) f(x)}{x-a}$ is

$\lim _{x \rightarrow \pi / 6} \left[ \frac{3 \sin x - \sqrt{3} \cos x}{6x - \pi} \right]$ is equal to:

$\mathop {\lim }\limits_{\theta \to \frac{\pi }{2}} \frac{\frac{\pi }{2} - \theta}{\cot \theta} =$

If $f(1) = 1$ and $f'(1) = 4,$ then the value of $\mathop {\lim }\limits_{x \to 1} \frac{{\sqrt {f(x)} - 1}}{{\sqrt x - 1}}$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module