Question

Evaluate: \( \lim_{x \to 0} \frac{\sqrt[3]{1+x} - \sqrt[3]{1-x}}{x} \)

• A. 1
• B. 0
• C. \( \frac{2}{3} \)
• D. \( \frac{1}{3} \)

Hint:

For expressions like \( f(x)-f(-x) \), symmetric expansion or derivative trick \( 2f'(0) \) is very useful.

Answer 1

The Correct Option is C

Step 1: Use binomial expansion.
For small \( x \):
\[ (1+x)^{1/3} \approx 1 + \frac{x}{3} \]
\[ (1-x)^{1/3} \approx 1 - \frac{x}{3} \]

Step 2: Substitute in numerator.
\[ \sqrt[3]{1+x} - \sqrt[3]{1-x} \approx \left(1 + \frac{x}{3}\right) - \left(1 - \frac{x}{3}\right) \]
\[ = \frac{x}{3} + \frac{x}{3} = \frac{2x}{3} \]

Step 3: Substitute into limit.
\[ \lim_{x \to 0} \frac{\frac{2x}{3}}{x} \]

Step 4: Cancel \( x \).
\[ = \frac{2}{3} \]

Step 5: Alternative method (derivative concept).
\[ \lim_{x \to 0} \frac{f(x) - f(-x)}{x} = 2f'(0) \]
Here \( f(x) = (1+x)^{1/3} \).

Step 6: Compute derivative.
\[ f'(x) = \frac{1}{3}(1+x)^{-2/3} \Rightarrow f'(0) = \frac{1}{3} \]
\[ \Rightarrow 2f'(0) = \frac{2}{3} \]

Step 7: Final conclusion.
\[ \boxed{\frac{2}{3}} \]