Question

\[ \lim_{y \to 0} \frac{\sqrt{1 + \sqrt{1 + y^4}} - \sqrt{2}}{y^4} = \ ? \]

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

Hint:

When evaluating limits involving nested square roots and small values, consider using binomial or Taylor approximations to simplify expressions.

Answer 1

The Correct Option is A

To solve the limit problem \(\lim_{y \to 0} \frac{\sqrt{1 + \sqrt{1 + y^4}} - \sqrt{2}}{y^4}\), we use a binomial expansion to approximate the square root functions when \(y\) is approaching zero. We have:

\[ \sqrt{1 + y^4} \approx 1 + \frac{y^4}{2} - \frac{y^8}{8} + \cdots \]

Substituting this into the outer square root, we get:

\[ \sqrt{1 + \sqrt{1 + y^4}} \approx \sqrt{1 + \left(1 + \frac{y^4}{2} - \frac{y^8}{8}\right)} \approx \sqrt{2 + \frac{y^4}{2} - \frac{y^8}{8}} \]

Expanding again using binomial approximation:

\[ \sqrt{2 + \frac{y^4}{2} - \frac{y^8}{8}} \approx \sqrt{2} + \frac{\frac{y^4}{2}}{2\sqrt{2}} = \sqrt{2} + \frac{y^4}{4\sqrt{2}} \]

Now, replace back into the original limit expression:

\[ \frac{\left(\sqrt{2} + \frac{y^4}{4\sqrt{2}}\right) - \sqrt{2}}{y^4} = \frac{\frac{y^4}{4\sqrt{2}}}{y^4} = \frac{1}{4\sqrt{2}} \]

Therefore, the limit evaluates to:

\[ \boxed{\frac{1}{4\sqrt{2}}} \]

The correct answer is \( \frac{1}{4\sqrt{2}} \).