Explanation:

To solve \(\lim_{x \to 1} \frac{x^3 - 1}{\sqrt{x} - 1}\):

- Substitute \(x = 1\): Both the numerator \(x^3 - 1\) and the denominator \(\sqrt{x} - 1\) become 0, leading to an indeterminate form \(0/0\).

- Factorize the Numerator:

- \(x^3 - 1\) can be factored using the difference of cubes formula: \((x - 1)(x^2 + x + 1)\).

- Reformulate the Expression:

- The expression becomes \(\frac{(x - 1)(x^2 + x + 1)}{\sqrt{x} - 1}\).

- Use Algebraic Manipulation (Rationalize):

- Multiply the numerator and the denominator by the conjugate: \(\sqrt{x} + 1\).

- This leads to: \(\frac{(x - 1)(x^2 + x + 1)(\sqrt{x} + 1)}{(x - 1)}\).

- Cancelling \(x - 1\), which results in \((x^2 + x + 1)(\sqrt{x} + 1)\).

- Evaluate the Limit:

- Substitute \(x = 1\) into this new expression:

- It becomes \((1^2 + 1 + 1)(\sqrt{1} + 1) = (3)(2) = 6\).

Answer: Option 3, 6