Question

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

• A. exists and is equal to $\frac{1}{2\sqrt{2}}$
• B. exists and is equal to $\frac{1}{4\sqrt{2}}$
• C. does not exist
• D. exists and is equal to $\frac{1}{2\sqrt{2}\left(\sqrt{2}+1\right)}$

Answer 1

Answer: (B)

To solve this question, we need to apply the concept of rationalization two times.
Given limit is $\underset{y\to 0}{\lim }\frac{\sqrt{1+\sqrt{1+y^4}}-\sqrt{2}}{y^4}$
$=\underset{y\to 0}{\lim }\frac{\sqrt{1+\sqrt{1+y^4}}-\sqrt{2}}{y^4}\times \frac{\sqrt{1+\sqrt{1+y^4}}+\sqrt{2}}{\sqrt{1+\sqrt{1+y^4}}+\sqrt{2}}$
$=\underset{y\to 0}{\lim }\frac{\left(\sqrt{1+y^4}-1\right)}{y^4\left(\sqrt{1+\sqrt{1+y^4}}+\sqrt{2}\right)}\times \frac{\left(\sqrt{1+y^4}+1\right)}{\left(\sqrt{1+y^4}+1\right)}$
$=\underset{y\to 0}{\lim }\frac{y^4}{y^4\left(\sqrt{1+\sqrt{1+y^4}}+\sqrt{2}\right)\left(\sqrt{1+y^4}+1\right)}$
$=\underset{y\to 0}{\lim }\frac{1}{\left(\sqrt{1+\sqrt{1+y^4}}+\sqrt{2}\right)\left(\sqrt{1+y^4}+1\right)}$
$=\frac{1}{4\sqrt{2}}$