lim _{y rightarrow 0} frac{sqrt{1+sqrt{1+y^4} | vedclass

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Question

(A) Let $L = \lim _{y \rightarrow 0} \frac{\sqrt{1+\sqrt{1+y^4}}-\sqrt{2}}{y^4}$.Rationalizing the numerator,we multiply by $\frac{\sqrt{1+\sqrt{1+y^4

Vedclass · Accepted Answer

$\frac{1}{4 \sqrt{2}}$

Vedclass · Answer

(A) Let $L = \lim _{y ightarrow 0} \frac{\sqrt{1+\sqrt{1+y^4}}-\sqrt{2}}{y^4}$.Rationalizing the numerator,we multiply by $\frac{\sqrt{1+\sqrt{1+y^4}}+\sqrt{2}}{\sqrt{1+\sqrt{1+y^4}}+\sqrt{2}}$:$L = \lim _{y ightarrow 0} \frac{(1+\sqrt{1+y^4})-2}{y^4(\sqrt{1+\sqrt{1+y^4}}+\sqrt{2})} = \lim _{y ightarrow 0} \frac{\sqrt{1+y^4}-1}{y^4(\sqrt{1+\sqrt{1+y^4}}+\sqrt{2})}$.Rationalizing the numerator again,we multiply by $\frac{\sqrt{1+y^4}+1}{\sqrt{1+y^4}+1}$:$L = \lim _{y ightarrow 0} \frac{(1+y^4)-1}{y^4(\sqrt{1+\sqrt{1+y^4}}+\sqrt{2})(\sqrt{1+y^4}+1)} = \lim _{y ightarrow 0} \frac{y^4}{y^4(\sqrt{1+\sqrt{1+y^4}}+\sqrt{2})(\sqrt{1+y^4}+1)}$.Canceling $y^4$ and evaluating the limit as $y ightarrow 0$:$L = \frac{1}{(\sqrt{1+1}+\sqrt{2})(\sqrt{1}+1)} = \frac{1}{(\sqrt{2}+\sqrt{2})(1+1)} = \frac{1}{2\sqrt{2} 	imes 2} = \frac{1}{4\sqrt{2}}$.

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

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