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We are given a piecewise function:

\[
f(x) =
\begin{cases}
x^2 - 2x, & x < 2 \\
1, & x = 2 \\
x^2 - 6x + 8, & x > 2
\end{cases}
\]

We are tasked with finding the **left-hand limit** of the function as \( x \to 2^- \).

### Step 1: Identify the relevant piece of the function for \( x \to 2^- \)
As \( x \to 2^- \), we approach 2 from the left, meaning we must use the part of the function where \( x < 2 \). This gives us the expression:
\[
f(x) = x^2 - 2x \quad \text{for} \quad x < 2.
\]

### Step 2: Evaluate the limit of \( f(x) = x^2 - 2x \) as \( x \to 2^- \)

Now, let's compute the limit:
\[
\lim_{x \to 2^-} (x^2 - 2x) = (2)^2 - 2(2) = 4 - 4 = 0.
\]

### Conclusion:
The left-hand limit as \( x \to 2^- \) is:
\[
\lim_{x \to 2^-} f(x) = 0.
\]

Let me know if you'd like more details or further questions!

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Here are 5 related questions for further exploration:

1. How do you determine the right-hand limit \( \lim_{x \to 2^+} f(x) \)?
2. What is the overall limit \( \lim_{x \to 2} f(x) \)?
3. How do you know if a piecewise function is continuous at a given point?
4. Can the limit exist even if the function is not defined at the point (e.g., \( x = 2 \))?
5. How do limits from both sides affect differentiability?

**Tip:** When evaluating limits of piecewise functions, always ensure you're selecting the correct piece of the function based on the direction of the approach (from left or right).

Determine the left-hand limit of the piecewise function as x approaches 2 from the left:

f(x) = { x^2 - 2x, for x < 2; 1, for x = 2; x^2 - 6x + 8, for x > 2. }

Find the left-hand limit of a piecewise function as x approaches 2 from the left. This problem explores limits, quadratic expressions, and piecewise functions, making it suitable for high school students in Grades 10-12.

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