ExamBro
ExamBro
KCET · Maths · Limits

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

  1. A \(\frac{1}{2 \sqrt{3}}\)
  2. B \(\frac{1}{3 \sqrt{2}}\)
  3. C \(2 \sqrt{3}\)
  4. D \(3 \sqrt{2}\)
Verified Solution

Answer & Solution

Correct Answer

(A) \(\frac{1}{2 \sqrt{3}}\)

Step-by-step Solution

Detailed explanation

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