The given limit problem is:

$\lim_{x \to 1} \frac{x - 1}{\sqrt{x} - \sqrt{1}}$

To solve this, we will use algebraic manipulation, particularly multiplying by the conjugate to simplify the denominator.

Step 1: Multiply by the conjugate

The conjugate of $\sqrt{x} - \sqrt{1}$ is $\sqrt{x} + \sqrt{1}$. We multiply both the numerator and the denominator by this conjugate:

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

Step 2: Simplify the denominator

Using the difference of squares formula:

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

So the expression becomes:

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

Step 3: Cancel out common factors

The factor $x - 1$ cancels from both the numerator and the denominator (as long as $x \neq 1$):

$\sqrt{x} + 1$

Step 4: Substitute the limit

Now, substitute $x = 1$:

$\sqrt{1} + 1 = 1 + 1 = 2$

Thus, the value of the limit is:

$\boxed{2}$

Let me know if you need further details or clarifications!

Follow-up questions:

1. What is the purpose of multiplying by the conjugate in limit problems?
2. Why does the factor $x - 1$ cancel out, and why is this valid for $x \neq 1$?
3. How would this limit change if we had a different value for the square roots?
4. Can L'Hôpital's Rule be applied to this limit problem? If so, how?
5. How do we handle limits when the expression inside the limit results in an indeterminate form like $0/0$?

Tip:

When faced with square roots in limits, multiplying by the conjugate is a powerful technique to simplify the expression.