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After examining the image, the problem appears to be structured as follows:

**Problem 2:**
We are asked to **show that** \(\lim_{x \to 2} f(x) = 4\) given certain conditions.

**Conditions:**
\[
f(x) = \begin{cases} 
    \frac{x^2 - 4}{x - 2} & \text{for } x \neq 2, \\
    1 & \text{for } x = 2.
\end{cases}
\]

### Solution Process:

#### Step 1: Simplifying the expression \(\frac{x^2 - 4}{x - 2}\)
\[
x^2 - 4 = (x - 2)(x + 2),
\]
so
\[
\frac{x^2 - 4}{x - 2} = \frac{(x - 2)(x + 2)}{x - 2}.
\]
For \(x \neq 2\), we can cancel \(x - 2\), which simplifies to:
\[
x + 2.
\]

#### Step 2: Taking the limit as \(x \to 2\)
\[
\lim_{x \to 2} f(x) = \lim_{x \to 2} (x + 2) = 4.
\]

Thus, we have shown that:
\[
\lim_{x \to 2} f(x) = 4.
\]

Would you like further clarification or more detailed steps on any part of this solution? 

Here are 5 related questions to deepen your understanding:

1. Why is it necessary to cancel the \((x - 2)\) term when simplifying the function?
2. What happens if we try to directly substitute \(x = 2\) into the original function?
3. What is the significance of defining \(f(x) = 1\) at \(x = 2\)?
4. How does this problem relate to the concept of removable discontinuities in functions?
5. What other limits involving piecewise functions could have a similar solution process?

**Tip:** When simplifying rational functions with a common factor, always check if it's safe to cancel the factor, especially when it's zero at certain points.

Show that \(\lim_{x \to 2} f(x) = 4\), where \(f(x)\) is defined as a piecewise function involving rational and constant expressions.

Understand how to evaluate the limit of a piecewise function as x approaches 2. The solution involves simplifying a rational expression through factoring. Suitable for students in Grades 11-12, covering concepts of limits and simplification.

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