You've uploaded a set of limit problems. I'll solve a few of them step by step to guide you through.

First, let’s start with this one:

$\lim_{{x \to 2}} \frac{x - 2}{\sqrt{x} - 2}$

This limit appears to lead to an indeterminate form $\frac{0}{0}$, so we can apply algebraic manipulation to solve it. Specifically, we rationalize the denominator.

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

Simplifying the expression:

$= \lim_{{x \to 2}} \frac{(x - 2)(\sqrt{x} + 2)}{(\sqrt{x})^2 - 2^2} = \lim_{{x \to 2}} \frac{(x - 2)(\sqrt{x} + 2)}{x - 4}$

Since $x - 2 = (x - 4 + 2)$, we can cancel out $x - 4$ and substitute $x = 2$:

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

Thus, the limit is: $\sqrt{2} + 2$

Let me know if you'd like more solutions from the image, details about this solution, or if you have any specific questions.

Here are 5 related questions to expand your understanding:

1. How would you apply L'Hopital's Rule to evaluate limits of indeterminate forms?
2. Can you generalize the method of rationalizing for limits involving square roots?
3. What happens if $x \to \infty$ in a limit involving square roots?
4. How do you evaluate limits when they approach values from the left versus the right?
5. How does multiplying by the conjugate help simplify complex limits?

Tip: Always check if the limit leads to an indeterminate form $\frac{0}{0}$ or $\infty/\infty$, as it may suggest L'Hopital's Rule or algebraic manipulation like rationalization is needed.