The expression in the image is:

$\lim_{x \to 5} \frac{x - 5}{\sqrt{x} - \sqrt{5}}$

To evaluate this limit as $x \to 5$, a direct substitution results in a $\frac{0}{0}$ indeterminate form, so we need to simplify the expression. Here's how we can proceed:

1. Multiply the numerator and denominator by the conjugate of the denominator, $\sqrt{x} + \sqrt{5}$, to eliminate the square roots: $\frac{x - 5}{\sqrt{x} - \sqrt{5}} \cdot \frac{\sqrt{x} + \sqrt{5}}{\sqrt{x} + \sqrt{5}} = \frac{(x - 5)(\sqrt{x} + \sqrt{5})}{(\sqrt{x})^2 - (\sqrt{5})^2}$
2. Simplifying the denominator using the difference of squares: $= \frac{(x - 5)(\sqrt{x} + \sqrt{5})}{x - 5}$
3. Cancel the common factor $x - 5$ (as long as $x \neq 5$): $= \sqrt{x} + \sqrt{5}$
4. Now, substitute $x = 5$ into the simplified expression: $\sqrt{5} + \sqrt{5} = 2\sqrt{5}$

Thus, the limit is: $\lim_{x \to 5} \frac{x - 5}{\sqrt{x} - \sqrt{5}} = 2\sqrt{5}$

Would you like more details on any of the steps?

Here are 5 related questions:

1. What are the common techniques for evaluating limits that result in indeterminate forms?
2. How does multiplying by the conjugate help simplify limits with square roots?
3. What is the difference of squares formula, and how is it used in limits?
4. What other types of indeterminate forms exist in limits, and how are they handled?
5. How does L'Hopital's rule apply to limits involving indeterminate forms?

Tip: Always check for simple factorizations or conjugates when dealing with indeterminate limits involving square roots.