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The given limit problem is:

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

### Step 1: Simplifying the limit expression
When \( x \to 0 \), both the numerator and denominator become indeterminate of the form \( 0/0 \). We can resolve this indeterminate form by applying the first terms of the binomial series expansion for \( \sqrt{x+1} \) and \( \sqrt[3]{x+1} \).

#### Expansion of \( \sqrt{x+1} \) around \( x = 0 \):
Using the binomial expansion:
\[
\sqrt{x+1} = 1 + \frac{x}{2} + O(x^2)
\]
For small \( x \), we approximate:
\[
\sqrt{x+1} \approx 1 + \frac{x}{2}
\]

#### Expansion of \( \sqrt[3]{x+1} \) around \( x = 0 \):
Similarly, using the binomial expansion:
\[
\sqrt[3]{x+1} = 1 + \frac{x}{3} + O(x^2)
\]
For small \( x \), we approximate:
\[
\sqrt[3]{x+1} \approx 1 + \frac{x}{3}
\]

### Step 2: Substituting the expansions
Now, substitute these approximations into the original limit:

\[
\frac{1 - \left(1 + \frac{x}{2}\right)}{1 - \left(1 + \frac{x}{3}\right)} = \frac{-\frac{x}{2}}{-\frac{x}{3}}
\]

### Step 3: Simplifying the expression
This simplifies to:
\[
\frac{-\frac{x}{2}}{-\frac{x}{3}} = \frac{3}{2}
\]

### Final Answer:
\[
\lim_{x \to 0} \left( \frac{1 - \sqrt{x+1}}{1 - \sqrt[3]{x+1}} \right) = \frac{3}{2}
\]

Let me know if you want more details on the steps. Here are some related questions:

1. What happens if we apply L'Hopital's Rule instead of using binomial expansions?
2. How can we extend this problem to limits involving other roots, like the fourth or fifth roots?
3. What if the limit was as \( x \to 1 \) instead of \( x \to 0 \)?
4. How does this approach change for higher-order terms in the expansions of \( \sqrt{x+1} \) and \( \sqrt[3]{x+1} \)?
5. Can this type of limit be solved using a Taylor series?

**Tip:** For indeterminate forms like \( 0/0 \), binomial expansions or L'Hopital's Rule are great tools to resolve the limit.

lim(x→0) (1 - √(x+1)) / (1 - ³√(x+1))

Explore how to solve the limit problem lim(x→0) (1 - √(x+1)) / (1 - ³√(x+1)) using binomial expansions. This advanced calculus problem involves understanding indeterminate forms and applying series expansions. Suitable for Grades 11-12 students.

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