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Q. $\displaystyle\lim_{y \to0} \frac{\sqrt{1+ \sqrt{1+y^{4}}} -\sqrt{2}}{y^{4}} $

JEE MainJEE Main 2019Limits and Derivatives

Solution:

$\lim_{y \to0} \frac{\sqrt{1+\sqrt{1+y^{4}} } - \sqrt{2}}{y^{4}} =\lim_{y\to0} \frac{1+\sqrt{1+y^{4}} -2}{y^{4} \left(\sqrt{1+\sqrt{1+y^{4}} } + \sqrt{2}\right)} $
$ = \lim_{y\to0} \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\to0} \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\to0} \frac{1}{\left(\sqrt{1+\sqrt{1+y^{4}}} +\sqrt{2}\right)\left( \sqrt{1+y^{4}} +1\right)} = \frac{1}{4\sqrt{2}} $