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

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

Solution

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

Asked in: MHT CET 2024 (16 May Shift 2)

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