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The image you uploaded contains the following mathematical expression:

\[
\lim_{{x \to 2}} \frac{{x^2 + 8}}{{x - 4}}
\]

Let's solve this step by step.

### Step 1: Analyze the Expression

The expression is a limit as \(x\) approaches 2 of the function \(\frac{x^2 + 8}{x - 4}\). The first step in solving this limit is to substitute \(x = 2\) into the expression to see if it gives us a defined value or if we need to simplify further.

Substitute \(x = 2\):

\[
\frac{(2)^2 + 8}{2 - 4} = \frac{4 + 8}{-2} = \frac{12}{-2} = -6
\]

### Step 2: Conclusion

Since substituting \(x = 2\) directly into the function gives us a defined value of \(-6\), this is the value of the limit.

\[
\lim_{{x \to 2}} \frac{x^2 + 8}{x - 4} = -6
\]

### Final Answer

The limit is \(-6\).

Would you like further details or have any questions?

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Here are 5 related questions that expand on this concept:

1. What happens if the limit as \(x\) approaches a different number, say \(x \to 4\), for this function?
2. How would the answer change if the numerator was \(x^2 - 4\) instead of \(x^2 + 8\)?
3. Can you find the limit if the function is changed to \(\lim_{{x \to 2}} \frac{x^2 + 8}{x - 2}\)?
4. What is the general approach to solving limits where direct substitution leads to an indeterminate form?
5. How can L'Hôpital's rule be applied to solve limits involving indeterminate forms like \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \)?

**Tip:** Always try substituting the value into the limit expression first. If it results in a defined number, that's your limit. If it results in an indeterminate form (like \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \)), you'll need to simplify or use a rule like L'Hôpital's.

Learn how to solve the limit of (x^2 + 8) / (x - 4) as x approaches 2 with a detailed step-by-step solution. This problem involves limits and rational functions, focusing on direct substitution and determining the value of the limit. Suitable for students in Grades 11-12.

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