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

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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} = $

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