Question

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

• A. $\frac{1}{4\sqrt{2}}$
• B. $\frac{1}{2\sqrt{2}}$
• C. $\frac{1}{\sqrt{2}}$
• D. does not exists

Answer 1

The correct option is A $\frac{1}{4\sqrt{2}}$

$\underset{y\to 0}{lim}\frac{\sqrt{1+\sqrt{1+y^4}}-\sqrt{2}}{y^4}$
Now, rationalizing the numerator, we get
$\underset{y\to 0}{lim}\frac{1+\sqrt{1+y^4}-2}{y^4}\times \frac{1}{\sqrt{1+\sqrt{1+y^4}}+\sqrt{2}}$
$=\underset{y\to 0}{lim}\frac{\sqrt{1+y^4}-1}{y^4}\times \frac{1}{2\sqrt{2}}$
$=\underset{y\to 0}{lim}\frac{\sqrt{1+y^4}-1}{y^4}\times \frac{1}{2\sqrt{2}}$
Applying L Hospital's rule
$=\frac{1}{2\sqrt{2}}\underset{y\to 0}{lim}\frac{\frac{4y^3}{2\sqrt{1+y^4}}}{4y^3}$
$=\frac{1}{2\sqrt{2}}\times \frac{1}{2}=\frac{1}{4\sqrt{2}}$