Answer
$\frac{1}{3}$
Work Step by Step
To solve the limit
\[
\lim_{{x \to 0}} \frac{\sqrt[3]{1 + x} - 1}{x}
\]
1. Check for Indeterminate Form: Substitute \( x = 0 \):
\[
\frac{\sqrt[3]{1 + 0} - 1}{0} = \frac{1 - 1}{0} = \frac{0}{0}
\]
Since this is an indeterminate form, we can apply L'Hôpital's Rule.
2. Differentiate the Numerator and Denominator:
- For the numerator \( f(x) = \sqrt[3]{1 + x} - 1 \), the derivative is:
\[
f'(x) = \frac{1}{3}(1 + x)^{-\frac{2}{3}}
\]
- For the denominator \( g(x) = x \), the derivative is:
\[
g'(x) = 1
\]
3. Apply L'Hôpital's Rule: Substitute the derivatives back into the limit:
\[
\lim_{{x \to 0}} \frac{f'(x)}{g'(x)} = \lim_{{x \to 0}} \frac{\frac{1}{3}(1 + x)^{-\frac{2}{3}}}{1}
\]
This simplifies to:
\[
\lim_{{x \to 0}} \frac{1}{3}(1 + x)^{-\frac{2}{3}} = \frac{1}{3}(1 + 0)^{-\frac{2}{3}} = \frac{1}{3}
\]
Thus, the solution to the limit is:
\[
\frac{1}{3}
\]
