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.