The mathematical expression in the image appears to be:

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

Let me solve this step-by-step:

Step 1: Check for indeterminate form

Substituting $x = 1$ directly: $\frac{\sqrt{1} - 1}{1 - 1} = \frac{0}{0}.$ This is an indeterminate form, so we need to simplify further.

Step 2: Simplify using algebraic manipulation

The numerator $\sqrt{x} - 1$ can be rationalized: $\frac{\sqrt{x} - 1}{x - 1} \cdot \frac{\sqrt{x} + 1}{\sqrt{x} + 1} = \frac{(\sqrt{x} - 1)(\sqrt{x} + 1)}{(x - 1)(\sqrt{x} + 1)}.$

In the numerator, we expand: $(\sqrt{x} - 1)(\sqrt{x} + 1) = x - 1.$

Thus, the expression simplifies to: $\frac{x - 1}{(x - 1)(\sqrt{x} + 1)} = \frac{1}{\sqrt{x} + 1}, \quad x \neq 1.$

Step 3: Evaluate the limit

Now substitute $x = 1$ into the simplified expression: $\frac{1}{\sqrt{1} + 1} = \frac{1}{1 + 1} = \frac{1}{2}.$

Final Answer:

$\lim_{x \to 1} \frac{\sqrt{x} - 1}{x - 1} = \frac{1}{2}.$

Let me know if you'd like more details or clarifications. Here are 5 related questions for practice:

1. How do you rationalize the numerator of a fraction involving square roots?
2. What are other methods to solve indeterminate forms like $\frac{0}{0}$?
3. How does the concept of limits apply to functions with radicals?
4. What is L'Hôpital's Rule, and could it be used here?
5. How can you check if a function is continuous at a given point?

Tip: When solving limits, always check if simplifying the expression or rationalizing helps before applying advanced techniques like L'Hôpital's Rule!