Question

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

• A. 0
• B. $\frac{1}{2\sqrt{2}}$
• C. $\frac{1}{4\sqrt{2}}$
• D. $\frac{1}{2\sqrt{2}(\sqrt{2}+1)}$

Answer 1

Answer: (C)

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

Answer 2

To evaluate the limit:

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

1. Approximate $\sqrt{1+y^4}$ for small $y$: $\sqrt{1+y^4} \approx 1 + \frac{y^4}{2}$

2. Substitute into the outer square root: $\sqrt{1+\sqrt{1+y^4}} \approx \sqrt{1 + (1+\frac{y^4}{2})} = \sqrt{2 + \frac{y^4}{2}}$

3. Use the linear approximation for the square root: $\sqrt{2+\epsilon} \approx \sqrt{2} + \frac{\epsilon}{2\sqrt{2}}$, where $\epsilon = \frac{y^4}{2}$. Thus, $\sqrt{2+\frac{y^4}{2}} \approx \sqrt{2} + \frac{\frac{y^4}{2}}{2\sqrt{2}} = \sqrt{2} + \frac{y^4}{4\sqrt{2}}$

4. Subtract $\sqrt{2}$ and divide by $y^4$: $\frac{\sqrt{1+\sqrt{1+y^4}}-\sqrt{2}}{y^4} \approx \frac{\frac{y^4}{4\sqrt{2}}}{y^4} = \frac{1}{4\sqrt{2}}$

Therefore, the limit is $\frac{1}{4\sqrt{2}}$.