The limit in the problem you've shared is:

$\lim_{x \to 2} \frac{x - 1}{\sqrt{x^2 + 3} - 2}$

We will use algebraic techniques to simplify and solve this limit. First, let's check what happens if we directly substitute $x = 2$.

Direct substitution:

$\frac{2 - 1}{\sqrt{2^2 + 3} - 2} = \frac{1}{\sqrt{4 + 3} - 2} = \frac{1}{\sqrt{7} - 2}$

This doesn't yield an indeterminate form like $\frac{0}{0}$ or $\frac{\infty}{\infty}$, so the limit can be calculated directly.

Thus, the answer is:

$\frac{1}{\sqrt{7} - 2}$

For further simplification, if you rationalize the denominator:

$\frac{1}{\sqrt{7} - 2} \times \frac{\sqrt{7} + 2}{\sqrt{7} + 2} = \frac{\sqrt{7} + 2}{(\sqrt{7})^2 - 2^2} = \frac{\sqrt{7} + 2}{7 - 4} = \frac{\sqrt{7} + 2}{3}$

So, the simplified result of the limit is:

$\frac{\sqrt{7} + 2}{3}$

Would you like more details on any step? Here are some related questions:

1. How do you simplify radicals in the denominator?
2. What is rationalizing the denominator, and why is it used?
3. Can L'Hôpital's rule be applied here? If so, how?
4. How do limits behave when direct substitution works?
5. What are indeterminate forms and when do they occur in limits?

Tip: Always check the form of the expression when taking limits to decide whether direct substitution, factoring, or rationalizing is appropriate.