Q82 / 180 · KCET 2022 · Mathematics

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

  • Correct answer
    A
    $\frac{1}{2 \sqrt{3}}$
  • B
    $\frac{1}{3 \sqrt{2}}$
  • C
    $2 \sqrt{3}$
  • D
    $3 \sqrt{2}$

Solution

Let $L=\lim _{y \rightarrow 0} \frac{\sqrt{3+y^3}-\sqrt{3}}{y^3}$ $$ \begin{aligned} L & =\lim _{y \rightarrow 0} \frac{\left(\sqrt{3+y^3}-\sqrt{3}\right)}{y^3} \times \frac{\left(\sqrt{3+y^3}+\sqrt{3}\right)}{\left(\sqrt{3+y^3}+\sqrt{3}\right)} \\ & =\lim _{y \rightarrow 0} \frac{3+y^3-3}{y^3\left(\sqrt{3+y^3}+\sqrt{3}\right)}=\lim _{y \rightarrow 0} \frac{y^3}{y^3\left(\sqrt{3+y^3}+\sqrt{3}\right)} \\ & =\lim _{y \rightarrow 0} \frac{1}{\sqrt{3+y^3}+\sqrt{3}}=\frac{1}{\sqrt{3}+\sqrt{3}}=\frac{1}{2 \sqrt{3}} \end{aligned} $$

Hence, the correct answer is option: 'A'.

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