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. exists and equals $\frac{1}{2\sqrt{2}}$
• D. exists and equals $\frac{1}{2\sqrt{2}(\sqrt{2}+1)}$

Answer 1

Answer: (A)

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