We need to evaluate the limit:

$\lim_{x \to 1} \frac{\sqrt{1 - x^3} - \sqrt{1 - x}}{\sqrt{1 - x^2}}$

Step 1: Analyze direct substitution

If we substitute $x = 1$ directly, all terms under the square root become 0, resulting in an indeterminate form $0/0$. Therefore, we need to simplify the expression further.

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Step 2: Rationalize the numerator

To simplify the numerator $\sqrt{1 - x^3} - \sqrt{1 - x}$, multiply and divide by its conjugate:

$\sqrt{1 - x^3} - \sqrt{1 - x} = \frac{\left(\sqrt{1 - x^3} - \sqrt{1 - x}\right)\left(\sqrt{1 - x^3} + \sqrt{1 - x}\right)}{\sqrt{1 - x^3} + \sqrt{1 - x}} = \frac{(1 - x^3) - (1 - x)}{\sqrt{1 - x^3} + \sqrt{1 - x}}$

This simplifies to:

$\frac{(1 - x^3) - (1 - x)}{\sqrt{1 - x^3} + \sqrt{1 - x}} = \frac{x - x^3}{\sqrt{1 - x^3} + \sqrt{1 - x}}$

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Step 3: Rewrite the original expression

Now the original limit becomes:

$\lim_{x \to 1} \frac{\frac{x - x^3}{\sqrt{1 - x^3} + \sqrt{1 - x}}}{\sqrt{1 - x^2}}$

This simplifies to:

$\lim_{x \to 1} \frac{x - x^3}{\sqrt{1 - x^3} + \sqrt{1 - x}} \cdot \frac{1}{\sqrt{1 - x^2}}$

$= \lim_{x \to 1} \frac{(x - x^3)}{\sqrt{1 - x^3} + \sqrt{1 - x}} \cdot \frac{1}{\sqrt{1 - x^2}}$

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Step 4: Factorize $x - x^3$

Factor $x - x^3$ as $x(1 - x^2)$. Substituting this gives:

$\lim_{x \to 1} \frac{x(1 - x^2)}{\left(\sqrt{1 - x^3} + \sqrt{1 - x}\right) \sqrt{1 - x^2}}$

Cancel $1 - x^2$ from the numerator and denominator (valid because $x \to 1$ and $1 - x^2 \neq 0$):

$\lim_{x \to 1} \frac{x}{\sqrt{1 - x^3} + \sqrt{1 - x}}$

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Step 5: Expand $1 - x^3$ using a Taylor approximation

As $x \to 1$, approximate $1 - x^3$ and $1 - x$:

• $1 - x^3 \approx 3(1 - x)$ (since $x^3 = 1 - 3(1 - x)$ to first order)
• Substitute this into $\sqrt{1 - x^3}$: $\sqrt{1 - x^3} \approx \sqrt{3(1 - x)}$

Thus, $\sqrt{1 - x^3} + \sqrt{1 - x} \approx \sqrt{3(1 - x)} + \sqrt{1 - x}$.

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Step 6: Simplify further

Factor out $\sqrt{1 - x}$ from $\sqrt{3(1 - x)} + \sqrt{1 - x}$:

$\sqrt{3(1 - x)} + \sqrt{1 - x} = \sqrt{1 - x} \left(\sqrt{3} + 1\right)$

Substitute back into the limit:

$\lim_{x \to 1} \frac{x}{\sqrt{1 - x} \cdot \left(\sqrt{3} + 1\right)}$

Since $x \to 1$, replace $x$ with 1:

$\frac{1}{\sqrt{3} + 1}$

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Final Answer:

$\boxed{\frac{1}{\sqrt{3} + 1}}$

If needed, this can be rationalized further:

$\frac{1}{\sqrt{3} + 1} \cdot \frac{\sqrt{3} - 1}{\sqrt{3} - 1} = \frac{\sqrt{3} - 1}{2}$

Alternate simplified form: $\boxed{\frac{\sqrt{3} - 1}{2}}$

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Do you have questions or need further clarification?
Here are related questions to explore:

1. How do you apply conjugates to simplify square root expressions in limits?
2. What is the Taylor approximation for functions like $1 - x^3$ around $x = 1$?
3. Why is rationalizing the numerator helpful in indeterminate limits?
4. How can you check the validity of cancellations in limits?
5. What are common indeterminate forms, and how do you resolve them?

Tip: Always analyze the dominant terms as $x$ approaches the limit point to simplify the computation!