Question

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

• A. $\frac{1}{4\sqrt{2}}$
• B. $\frac{1}{2\sqrt{2}}$
• C. $\frac{1}{2\sqrt{2}(1+\sqrt{2})}$
• D. Does not exist.

Answer 1

The correct option is A

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

Simplifying the given function, we get.

$$\begin{gathered} \underset{y\to 0}{\lim }\frac{\sqrt{1+\sqrt{1+y^4}}-\sqrt{2}}{y^4} \\ =\underset{y\to 0}{\lim }\frac{\sqrt{1+\sqrt{1+y^4}}-\sqrt{2}}{y^4}\times \frac{\sqrt{1+\sqrt{1+y^4}}+\sqrt{2}}{\sqrt{1+\sqrt{1+y^4}}+\sqrt{2}} \\ =\underset{y\to 0}{\lim }\frac{\sqrt{1+y^4}-1}{y^4\sqrt{1+\sqrt{1+y^4}}+\sqrt{2}} \\ =\underset{y\to 0}{\lim }\frac{\sqrt{1+y^4}-1}{y^4\sqrt{1+\sqrt{1+y^4}}+\sqrt{2}}\times \frac{\sqrt{1+y^4}+1}{\sqrt{1+y^4}+1} \\ =\underset{y\to 0}{\lim }\frac{y^4}{(y^4\sqrt{1+\sqrt{1+y^4}}+\sqrt{2})(\sqrt{1+y^4}+1)} \\ =\underset{y\to 0}{\lim }\frac{1}{\left(\sqrt{1+\sqrt{1+y^4}}+\sqrt{2}\right)\left(\sqrt{1+y^4}+1\right)} \\ =\frac{1}{4\sqrt{2}} \end{gathered}$$

Therefore, Option (A) is the correct answer.