Question

Evaluate the limit: \[ \lim_{x \to 0} \frac{\sqrt{1 + \sqrt{1 + x^4}} - \sqrt{2 + x^5 + x^6}}{x^4} =\]

• A. \( \frac{1}{4\sqrt{2}} \)
• B. \( \frac{1}{2\sqrt{2}} \)
• C. \( \frac{1}{\sqrt{2}} \)
• D. \( \frac{1}{3\sqrt{2}} \)

Hint:

For limits involving nested radicals, use the binomial approximation \( \sqrt{1 + x} \approx 1 + \frac{x}{2} \) for small \( x \) to simplify calculations.

Answer 1

The Correct Option is A

Step 1: Approximate the Square Root Expansions 
Using the first-order binomial approximation: \[ \sqrt{1 + x} \approx 1 + \frac{x}{2} \text{ for small } x. \] Expanding \( \sqrt{1 + x^4} \): \[ \sqrt{1 + x^4} \approx 1 + \frac{x^4}{2}. \] Thus, expanding the nested square root term: \[ \sqrt{1 + \sqrt{1 + x^4}} = \sqrt{1 + \left(1 + \frac{x^4}{2} - 1\right)} = \sqrt{1 + \frac{x^4}{2}}. \] Applying binomial approximation again: \[ \sqrt{1 + \frac{x^4}{2}} \approx 1 + \frac{x^4}{4}. \] 
Step 2: Approximate the Second Square Root Term 
Expanding \( \sqrt{2 + x^5 + x^6} \): \[ \sqrt{2 + x^5 + x^6} \approx \sqrt{2} \cdot \sqrt{1 + \frac{x^5}{2} + \frac{x^6}{2}}. \] Using the binomial expansion: \[ \sqrt{1 + \frac{x^5}{2} + \frac{x^6}{2}} \approx 1 + \frac{x^5}{4} + \frac{x^6}{4}. \] Thus, \[ \sqrt{2 + x^5 + x^6} \approx \sqrt{2} \left(1 + \frac{x^5}{4} + \frac{x^6}{4} \right) = \sqrt{2} + \frac{\sqrt{2} x^5}{4} + \frac{\sqrt{2} x^6}{4}. \] 
Step 3: Compute the Limit 
Now, the numerator simplifies to: \[ \left(1 + \frac{x^4}{4}\right) - \left(\sqrt{2} + \frac{\sqrt{2} x^5}{4} + \frac{\sqrt{2} x^6}{4}\right). \] Rearranging: \[ 1 - \sqrt{2} + \frac{x^4}{4} - \frac{\sqrt{2} x^5}{4} - \frac{\sqrt{2} x^6}{4}. \] For small \( x \), the dominant term in the numerator is: \[ \frac{x^4}{4}. \] Thus, the limit evaluates to: \[ \lim_{x \to 0} \frac{\frac{x^4}{4}}{x^4} = \frac{1}{4\sqrt{2}}. \] 
Final Answer: \( \boxed{\frac{1}{4\sqrt{2}}} \)

Answer 2

To solve the problem, we need to evaluate the limit:

\[\lim_{x \to 0} \frac{\sqrt{1 + \sqrt{1 + x^4}} - \sqrt{2 + x^5 + x^6}}{x^4}\]

Start by expanding both square root terms using a Taylor series expansion around \(x=0\).

The innermost expression \(\sqrt{1+x^4}\) can be expanded as:

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

Plug this into the outer square root:

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

Expanding this:

\[\approx \sqrt{2} + \frac{x^4}{4\sqrt{2}} + \cdots\]

Next, expand \(\sqrt{2+x^5+x^6}\):

\[\sqrt{2+x^5+x^6} \approx \sqrt{2} + \frac{x^5}{4\sqrt{2}} + \frac{x^6}{4\sqrt{2}} + \cdots\]

Substitute these expansions back into the original limit:

\[\lim_{x \to 0} \frac{\left(\sqrt{2} + \frac{x^4}{4\sqrt{2}}\right) - \left(\sqrt{2} + \frac{x^5}{4\sqrt{2}} + \frac{x^6}{4\sqrt{2}}\right)}{x^4}\]

Simplifying the expression inside the limit:

\[\lim_{x \to 0} \frac{\frac{x^4}{4\sqrt{2}} - \frac{x^5}{4\sqrt{2}} - \frac{x^6}{4\sqrt{2}}}{x^4}\]

Factoring out the common factor:

\[\lim_{x \to 0} \frac{x^4}{x^4} \cdot \frac{1}{4\sqrt{2}} \cdot (1-x-\frac{x^2}{4})\]

Evaluating the limit as \(x\) approaches zero:

\[= \frac{1}{4\sqrt{2}} \cdot (1-0-0) = \frac{1}{4\sqrt{2}}\]

Thus, the correct option is:

\( \frac{1}{4\sqrt{2}} \)