Q66 / 150 · MHT CET 2024 (16 May Shift 2) · Mathematics

$\lim _{y \rightarrow 0} \frac{\sqrt{1+\sqrt{1+y^4}}-\sqrt{2}}{y^4}=$

  • A
    0
  • B
    $\frac{1}{2 \sqrt{2}}$
  • Correct answer
    C
    $\frac{1}{4 \sqrt{2}}$
  • D
    $\frac{1}{2 \sqrt{2}(\sqrt{2}+1)}$

Solution

By rationalising, we get $\begin{aligned} & \lim _{y \rightarrow 0} \frac{\sqrt{1+\sqrt{1+y^4}-\sqrt{2}}}{y^4} \\ & =\lim _{y \rightarrow 0} \frac{1+\sqrt{1+y^4}-2}{y^4\left(\sqrt{1+\sqrt{1+y^4}}+\sqrt{2}\right)} \\ & =\lim _{y \rightarrow 0} \frac{\sqrt{1+y^4}-1}{y^4\left(\sqrt{1+\sqrt{1+y^4}}+\sqrt{2}\right)} \\ & =\lim _{y \rightarrow 0} \frac{\sqrt{1+y^4}-1}{y^4\left(\sqrt{1+\sqrt{1+y^4}}+\sqrt{2}\right)} \times \frac{\sqrt{1+y^4}+1}{\sqrt{1+y^4}+1} \\ & =\lim _{y \rightarrow 0} \frac{y^4}{y^4\left(\sqrt{1+\sqrt{1+y^4}}+\sqrt{2}\right)\left(\sqrt{1+y^4}+1\right)} \\ & =\frac{1}{(\sqrt{1+\sqrt{1+0}+\sqrt{2})(\sqrt{1+0}+1)}}=\frac{1}{4 \sqrt{2}} \end{aligned}$

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

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