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The given problem involves evaluating the limit:

\[
\lim_{x \to 1} \frac{\sqrt{x} - 1}{1 - \sqrt[3]{x}}
\]

To solve this, we follow these steps:

---

### Step 1: Recognize the indeterminate form
Substituting \(x = 1\):
\[
\frac{\sqrt{1} - 1}{1 - \sqrt[3]{1}} = \frac{0}{0},
\]
an indeterminate form. Thus, we need further simplification.

---

### Step 2: Rationalize numerator and denominator
#### Numerator:
Rationalize \(\sqrt{x} - 1\) by multiplying and dividing by \(\sqrt{x} + 1\):
\[
\sqrt{x} - 1 = \frac{(\sqrt{x} - 1)(\sqrt{x} + 1)}{\sqrt{x} + 1} = \frac{x - 1}{\sqrt{x} + 1}.
\]

#### Denominator:
Rationalize \(1 - \sqrt[3]{x}\) by using the difference of cubes:
\[
1 - \sqrt[3]{x} = \frac{(1 - \sqrt[3]{x})(1 + \sqrt[3]{x} + (\sqrt[3]{x})^2)}{1 + \sqrt[3]{x} + (\sqrt[3]{x})^2} = \frac{1 - x}{1 + \sqrt[3]{x} + (\sqrt[3]{x})^2}.
\]

---

### Step 3: Rewrite the limit
Now substitute these expressions back into the original limit:
\[
\lim_{x \to 1} \frac{\sqrt{x} - 1}{1 - \sqrt[3]{x}} = \lim_{x \to 1} \frac{\frac{x - 1}{\sqrt{x} + 1}}{\frac{1 - x}{1 + \sqrt[3]{x} + (\sqrt[3]{x})^2}}.
\]

Simplify by inverting the denominator and multiplying:
\[
\lim_{x \to 1} \frac{\sqrt{x} - 1}{1 - \sqrt[3]{x}} = \lim_{x \to 1} \frac{x - 1}{\sqrt{x} + 1} \cdot \frac{1 + \sqrt[3]{x} + (\sqrt[3]{x})^2}{1 - x}.
\]

Cancel \(x - 1\) and \(1 - x\) (note that \(1 - x = -(x - 1)\)):
\[
\lim_{x \to 1} \frac{\sqrt{x} - 1}{1 - \sqrt[3]{x}} = \lim_{x \to 1} \frac{-(1 + \sqrt[3]{x} + (\sqrt[3]{x})^2)}{\sqrt{x} + 1}.
\]

---

### Step 4: Substitute \(x = 1\)
\[
\frac{-(1 + \sqrt[3]{1} + (\sqrt[3]{1})^2)}{\sqrt{1} + 1} = \frac{-(1 + 1 + 1)}{1 + 1} = \frac{-3}{2}.
\]

---

### Final Answer:
\[
\lim_{x \to 1} \frac{\sqrt{x} - 1}{1 - \sqrt[3]{x}} = \boxed{-\frac{3}{2}}.
\]

Would you like a detailed explanation of any step?

---

#### Questions:
1. Why is rationalization used for simplifying limits?
2. How does the difference of cubes formula help in solving this problem?
3. Can L'Hôpital's rule be used to solve this limit instead?
4. What are the applications of such limits in calculus?
5. How would the approach change if the limit were multivariable?

#### Tip:
When solving limits involving roots, always check for indeterminate forms and consider rationalization or L’Hôpital’s rule to simplify calculations.

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

Discover how to solve the limit \(\lim_{x \to 1} \frac{\sqrt{x} - 1}{1 - \sqrt[3]{x}}\) step-by-step. This solution involves rationalization, the difference of cubes, and limit properties. Suitable for students learning introductory calculus concepts.

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