Question

The value of $\underset{y \rightarrow 0}{l i m}\frac{\sqrt{1 + \sqrt{1 + y^{4}}} - \sqrt{2}}{y^{4}}$

• 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 $\underset{y \rightarrow 0}{l i m}\frac{\sqrt{1 + \sqrt{1 + y^{4}}} - \sqrt{2}}{y^{4}}$
$=\underset{y \rightarrow 0}{l i m}\frac{\sqrt{1 + \sqrt{1 + y^{4}}} - \sqrt{2}}{y^{4}}\times \frac{\sqrt{1 + \sqrt{1 + y^{4}}} + \sqrt{2}}{\sqrt{1 + \sqrt{1 + y^{4}}} + \sqrt{2}}$
$=\underset{y \rightarrow 0}{l i m}\frac{\left(\sqrt{1 + y^{4}} - 1\right)}{y^{4} \left(\sqrt{1 + \sqrt{1 + y^{4}}} + \sqrt{2}\right)}\times \frac{\left(\sqrt{1 + y^{4}} + 1\right)}{\left(\sqrt{1 + y^{4}} + 1\right)}$
$=\underset{y \rightarrow 0}{l i m}\frac{y^{4}}{y^{4} \left(\sqrt{1 + \sqrt{1 + y^{4}}} + \sqrt{2}\right) \left(\sqrt{1 + y^{4}} + 1\right)}$
$=\underset{y \rightarrow 0}{l i m}\frac{1}{\left(\sqrt{1 + \sqrt{1 + y^{4}} \, } + \sqrt{2}\right) \left(\sqrt{1 + y^{4}} + 1\right)}$
$=\frac{1}{4 \sqrt{2}}$