Limit x tends to 0 of the cube root of (1+c*x)-1

We are asked to find the limit:

\[
\lim_{{x \to 0}} \left( \sqrt[3]{{1 + c \cdot x}} - 1 \right)
\]

To solve this, let's apply the technique of Taylor expansion around \( x = 0 \).

### Step-by-step solution:

The cube root function \( f(x) = \sqrt[3]{1 + x} \) can be expanded using the Taylor series around \( x = 0 \):

\[
\sqrt[3]{1 + x} = 1 + \frac{x}{3} - \frac{x^2}{9} + \mathcal{O}(x^3)
\]

Thus, substituting \( c \cdot x \) for \( x \):

\[
\sqrt[3]{1 + c \cdot x} = 1 + \frac{c \cdot x}{3} - \frac{(c \cdot x)^2}{9} + \mathcal{O}(x^3)
\]

Now, subtract 1 from both sides:

\[
\sqrt[3]{1 + c \cdot x} - 1 = \frac{c \cdot x}{3} - \frac{(c \cdot x)^2}{9} + \mathcal{O}(x^3)
\]

Taking the limit as \( x \to 0 \), the higher-order terms (\( x^2 \) and beyond) vanish, leaving us with:

\[
\lim_{{x \to 0}} \left( \frac{c \cdot x}{3} \right) = 0
\]

Thus, the limit is:

\[
\boxed{0}
\]

Would you like more details on any step or have further questions?

### Relative questions:
1. What is the Taylor series expansion of \( \sqrt[3]{1 + x} \) used for other cases?
2. How does the limit change for higher powers of \( x \) inside the cube root?
3. What is L'Hopital's Rule, and can it be applied to this limit?
4. Can this limit be solved using binomial expansion instead of Taylor series?
5. How does this limit behave if \( c \) is negative or large?

### Tip:
For small \( x \), Taylor series expansion is a powerful tool for approximating complex functions and solving limits.

Limit x tends to 0 of the cube root of (1 + c * x) - 1

Find the limit of (sqrt[3]{1 + c * x} - 1) as x approaches 0, using Taylor series expansion. This problem explores limits, series expansions, and their applications in calculus. Ideal for college-level calculus students.

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