We are tasked with finding the limit of $f(x)$ as $x$ approaches 2 from the left-hand side, i.e., $\lim_{x \to 2^-} f(x)$.

Key points to analyze:

• Left Curve (Region 1): This is the part of the graph that represents $f(x)$ for values of $x$ less than 2.
• The equation of the left curve is given by $f(x) = \frac{3}{4x - 2} + 4$.
• Right Curve (Region 2): The right side of the graph does not apply in this specific limit because we are interested in the behavior as $x \to 2^-$, which means approaching 2 from the left side.

Step-by-step process:

1. Equation of the left curve: $f(x) = \frac{3}{4x - 2} + 4$ As $x \to 2^-$, we substitute $x = 2$ in the function to analyze the behavior.

2. Substitute $x = 2$ into the expression: $4x - 2 = 4(2) - 2 = 8 - 2 = 6$ Therefore, the function becomes: $f(2^-) = \frac{3}{6} + 4 = 0.5 + 4 = 4.5$

Thus, the limit of $f(x)$ as $x$ approaches 2 from the left is: $\lim_{x \to 2^-} f(x) = 4.5$

Do you want details on why I evaluated this way or have any other questions?

Here are 5 related questions:

1. What is the difference between left-hand and right-hand limits?
2. How does the presence of an asymptote affect limits?
3. What would $\lim_{x \to 2^+} f(x)$ be for this function?
4. How do piecewise functions behave at points of discontinuity?
5. What role do open and closed circles play on a graph?

Tip: Always check whether a graph has discontinuities or jumps when evaluating limits near critical points!