lim _{y rightarrow 0} frac{sqrt{3+y^3}-sqrt{3 | vedclass

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Question

(A) Let $L = \lim _{y \rightarrow 0} \frac{\sqrt{3+y^3}-\sqrt{3}}{y^3}$ Rationalizing the numerator: $L = \lim _{y}$ ${\rightarrow 0} \frac{(\sqrt{3+y

Vedclass · Accepted Answer

$\frac{1}{2 \sqrt{3}}$

Vedclass · Answer

(A) Let $L = \lim _{y ightarrow 0} \frac{\sqrt{3+y^3}-\sqrt{3}}{y^3}$ Rationalizing the numerator: $L = \lim _{y}$ ${ightarrow 0} \frac{(\sqrt{3+y^3}-\sqrt{3})}{y^3} 	imes \frac{(\sqrt{3+y^3}+\sqrt{3})}{(\sqrt{3+y^3}+\sqrt{3})}$ $L = \lim _{y ightarrow 0} \frac{3+y^3-3}{y^3(\sqrt{3+y^3}+\sqrt{3})}$ $L = \lim _{y ightarrow 0} \frac{y^3}{y^3(\sqrt{3+y^3}+\sqrt{3})}$ $L = \lim _{y ightarrow 0} \frac{1}{\sqrt{3+y^3}+\sqrt{3}}$ Substituting $y = 0$: $L = \frac{1}{\sqrt{3+0}+\sqrt{3}} = \frac{1}{\sqrt{3}+\sqrt{3}} = \frac{1}{2 \sqrt{3}}$

$\lim _{y \rightarrow 0} \frac{\sqrt{3+y^3}-\sqrt{3}}{y^3} = $

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