Question

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

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

Answer 1

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

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

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

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

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

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