https://file.aiquickdraw.com/mathbot/file/1726044378424ii59dbz2.jpg

Let's go through the one-sided limits in the image:

### Problem 1:
\[
\lim_{{x \to 2^-}} \frac{3}{{x - 2}}
\]

Since this is a one-sided limit as \( x \to 2^- \), meaning \( x \) approaches 2 from the left, let's analyze the behavior of the function.

- As \( x \to 2^- \), the denominator \( x - 2 \) becomes slightly negative (since \( x \) is slightly less than 2).
- The numerator stays constant at 3.

Thus:
\[
\frac{3}{{x - 2}} \to -\infty \quad \text{as} \quad x \to 2^-
\]
Therefore:
\[
\lim_{{x \to 2^-}} \frac{3}{{x - 2}} = -\infty
\]

### Problem 2:
\[
\lim_{{x \to 9^-}} \frac{{\sqrt{x} - 3}}{{x - 9}}
\]

As \( x \to 9^- \), this also means approaching 9 from the left. We will apply factoring techniques for this.

First, notice that \( \sqrt{x} - 3 \) can be related to \( x - 9 \) because \( \sqrt{x} = 3 \) when \( x = 9 \). Factor this expression using a difference of squares:
\[
\frac{{\sqrt{x} - 3}}{{x - 9}} = \frac{{(\sqrt{x} - 3)(\sqrt{x} + 3)}}{{(x - 9)(\sqrt{x} + 3)}} = \frac{1}{{\sqrt{x} + 3}}
\]

Now, as \( x \to 9^- \), \( \sqrt{x} \) approaches 3, making the denominator approach 6:
\[
\lim_{{x \to 9^-}} \frac{1}{{\sqrt{x} + 3}} = \frac{1}{6}
\]

Thus, the limit is:
\[
\lim_{{x \to 9^-}} \frac{{\sqrt{x} - 3}}{{x - 9}} = \frac{1}{6}
\]

### Problem 3:
\[
\lim_{{x \to 2}} \frac{1}{{x^2 - 4}}
\]

Notice that \( x^2 - 4 \) can be factored as \( (x - 2)(x + 2) \). Thus:
\[
\lim_{{x \to 2}} \frac{1}{{x^2 - 4}} = \lim_{{x \to 2}} \frac{1}{{(x - 2)(x + 2)}}
\]

Now, as \( x \to 2 \), the term \( x - 2 \) goes to 0, while \( x + 2 \to 4 \), so the whole expression behaves like:
\[
\frac{1}{{0 \cdot 4}} = \infty
\]
Thus, the limit is infinite. However, because we are not explicitly given a one-sided limit, we can conclude:
\[
\lim_{{x \to 2}} \frac{1}{{x^2 - 4}} = \infty
\]

### Final Solutions:
1. \( \lim_{{x \to 2^-}} \frac{3}{{x - 2}} = -\infty \)
2. \( \lim_{{x \to 9^-}} \frac{{\sqrt{x} - 3}}{{x - 9}} = \frac{1}{6} \)
3. \( \lim_{{x \to 2}} \frac{1}{{x^2 - 4}} = \infty \)

Would you like more details or clarifications on any of these? 

Here are some related questions:
1. How do limits behave when approaching infinity?
2. What happens to limits at points where the denominator becomes 0?
3. How can the difference of squares be used in limit problems?
4. What are the implications of approaching limits from the left vs. right?
5. How does factoring help simplify limit expressions?

**Tip:** When evaluating limits with indeterminate forms like \( \frac{0}{0} \), try factoring or using L'Hopital's Rule.

Explore the complete solutions for one-sided limits, including the analysis of rational functions and factoring techniques. This guide covers important limit laws, behavior near points of discontinuity, and the application of difference of squares for simplifying expressions.

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