Question

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

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

Answer 1

The correct option is C exists and equals $\frac{1}{4\sqrt{2}}$

$\underset{y\to 0}{lim}\frac{\sqrt{1+\sqrt{1+y^4}}-\sqrt{2}}{y^4}$
Now rationalizing the numerator, we get
$\underset{y\to 0}{lim}\frac{1+\sqrt{1+y^4}-2}{y^4}\times \frac{1}{\sqrt{1+\sqrt{1+y^4}}+\sqrt{2}}$

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

$=\underset{y\to 0}{lim}\frac{\sqrt{1+y^4}-1}{y^4}\times \frac{1}{2\sqrt{2}}$
$=\frac{1}{2\sqrt{2}}\times \underset{y\to 0}{lim}\frac{\frac{4y^3}{2\sqrt{1+y^4}}}{4y^3}$
[Using L Hopital rule]

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