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Question:
$\lim _{y \rightarrow 0} \frac{\sqrt{1+\sqrt{1+y^{4}}}-\sqrt{2}}{y^{4}}$
Correct Option: 1
Solution:
$\lim _{y \rightarrow 0} \frac{\sqrt{1+\sqrt{1+y^{4}}}-\sqrt{2}}{y^{4}}$
$=\lim _{y \rightarrow 0} \frac{1+\sqrt{1+y^{4}}-2}{y^{4}\left(\sqrt{1+\sqrt{1+y^{4}}}+\sqrt{2}\right)}$
$=\lim _{y \rightarrow 0} \frac{\left(\sqrt{1+y^{4}}-1\right)\left(\sqrt{1+y^{4}}+1\right)}{y^{4}\left(\sqrt{1+\sqrt{1+y^{4}}}+\sqrt{2}\right)\left(\sqrt{1+y^{4}}+1\right)}$
$=\lim _{y \rightarrow 0} \frac{1+y^{4}-1}{y^{4}\left(\sqrt{1+\sqrt{1+y^{4}}}+\sqrt{2}\right)\left(\sqrt{1+y^{4}}+1\right)}$
$=\lim _{y \rightarrow 0} \frac{1}{\left(\sqrt{1+\sqrt{1+y^{4}}}+\sqrt{2}\right)\left(\sqrt{1+y^{4}}+1\right)}=\frac{1}{4 \sqrt{2}}$
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