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The image presents a problem asking to find three different limits. Let's go through each one step by step:

### Given expressions:
\[ 
\text{a) } \lim_{{x \to 2^+}} \frac{1}{{x^2 - 4}} 
\]
\[ 
\text{b) } \lim_{{x \to 2^-}} \frac{1}{{x^2 - 4}} 
\]
\[ 
\text{c) } \lim_{{x \to 2}} \frac{1}{{x^2 - 4}} 
\]

---

### Factoring \(x^2 - 4\):
\[
x^2 - 4 = (x - 2)(x + 2)
\]

This will help us analyze the limits since we know \(x = 2\) is a point where the denominator becomes zero, causing potential discontinuity.

---

### Limit a: \( \lim_{{x \to 2^+}} \frac{1}{{x^2 - 4}} \)

When \(x\) approaches \(2\) from the right (\(2^+\)), the expression in the denominator \(x^2 - 4 = (x - 2)(x + 2)\) becomes very small and positive because:
- \(x - 2\) is small and positive for \(x \to 2^+\).
- \(x + 2\) approaches 4 as \(x \to 2^+\).

Thus, the overall denominator is a very small positive number, which makes the fraction very large and positive.

**Conclusion**:
\[
\lim_{{x \to 2^+}} \frac{1}{{x^2 - 4}} = +\infty
\]

---

### Limit b: \( \lim_{{x \to 2^-}} \frac{1}{{x^2 - 4}} \)

When \(x\) approaches \(2\) from the left (\(2^-\)), the expression in the denominator \(x^2 - 4 = (x - 2)(x + 2)\) becomes very small and negative because:
- \(x - 2\) is small and negative for \(x \to 2^-\).
- \(x + 2\) is still approximately 4.

Thus, the overall denominator is a very small negative number, which makes the fraction very large but negative.

**Conclusion**:
\[
\lim_{{x \to 2^-}} \frac{1}{{x^2 - 4}} = -\infty
\]

---

### Limit c: \( \lim_{{x \to 2}} \frac{1}{{x^2 - 4}} \)

Since the limit as \(x\) approaches 2 from the left (\(2^-\)) and from the right (\(2^+\)) yield different results (\(-\infty\) and \(+\infty\), respectively), the two-sided limit does not exist.

**Conclusion**:
\[
\lim_{{x \to 2}} \frac{1}{{x^2 - 4}} \text{ does not exist.}
\]

---

Would you like more details or have any questions on this explanation?

Here are some related questions:
1. What does it mean for a limit to approach \( +\infty \)?
2. Why does factoring the expression help in solving the limits?
3. How can we visualize the behavior of a function as it approaches a vertical asymptote?
4. What are one-sided limits, and why are they important in discontinuities?
5. What are other techniques to handle indeterminate forms in limits?

**Tip**: Always check for factorization or simplifications in expressions, especially around points of discontinuity or where the denominator approaches zero.

Find the following limits: 
 a) lim x→2+ (1 / (x² - 4)) 
 b) lim x→2- (1 / (x² - 4)) 
 c) lim x→2 (1 / (x² - 4))

Learn how to solve the limit problems involving rational functions with discontinuities. This problem explores one-sided limits and behavior near vertical asymptotes using algebraic techniques such as factoring.

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