Question
The table below gives selected values for a function \( f \). Also shown is a portion of the graph of \( f \). The graph consists of a line segment for \( x < 3 \) and part of a parabola for \( x > 3 \). What is \( \lim_{x \rightarrow 3} f(x) \)?
\( x \) | 2.9 | 2.95 | 2.99 | 2.998 | 3.002 | 3.01 | 3.05 | 3.1 |
---|---|---|---|---|---|---|---|---|
\( f(x) \) | 1.650 | 1.625 | 1.605 | 1.601 | 1.602 | 1.612 | 1.659 | 1.716 |

A) 1.6
B) \( \frac{1.601 + 1.602}{2} \)
C) 2
D) The limit does not exist
▶️ Answer/Explanation
Solution
1. Graph Analysis: The graph shows continuity at \( x = 3 \) with values approaching approximately 1.6 from both sides.
2. Table Analysis: As \( x \) approaches 3 from both sides (2.998 → 3.002), \( f(x) \) approaches ≈1.6.
3. Linear Segment Calculation:
- Slope \( m = \frac{1.625 – 1.650}{2.95 – 2.9} = -0.5 \)
- Equation: \( y = -0.5(x – 2.9) + 1.65 \)
- At \( x = 3 \): \( y = -0.5(0.1) + 1.65 = 1.6 \)
4. Conclusion: The limit exists and equals 1.6.
✅ Answer: A) 1.6
Question
Let \( f \) be the piecewise function defined below. Also shown is a portion of the graph of \( f \). What is the value of \( \lim_{x \rightarrow 2} f(f(x)) \)?
\( f(x) = \begin{cases} -x^2 + 3x + 3 & \text{for } x < 2 \\ 6 & \text{for } x = 2 \\ 6 – \frac{1}{2}x & \text{for } x > 2 \end{cases} \)

A) -15
B) -7
C) 3
D) \( \frac{7}{2} \)
▶️ Answer/Explanation
Solution
1. Evaluate \( \lim_{x \to 2} f(x) \):
- Left limit (\( x \to 2^- \)): Use \(-x^2 + 3x + 3\) → \(-4 + 6 + 3 = 5\)
- Right limit (\( x \to 2^+ \)): Use \(6 – \frac{1}{2}x\) → \(6 – 1 = 5\)
- Since both limits equal 5, \( \lim_{x \to 2} f(x) = 5 \).
2. Compute \( f(f(x)) \) as \( x \to 2 \):
- Since \( f(x) \to 5 \) as \( x \to 2 \), evaluate \( f(5) \):
- For \( x = 5 > 2 \), use \( 6 – \frac{1}{2}x \) → \( 6 – 2.5 = 3.5 = \frac{7}{2} \).
3. Conclusion:
\[ \lim_{x \rightarrow 2} f(f(x)) = f(5) = \frac{7}{2} \]
✅ Answer: D) \( \frac{7}{2} \)
Question
The table above gives selected values for a function \( f \). Based on the data in the table, which of the following could not be the graph of \( f \) on the interval \( 2.9 \leq x \leq 3.1 \)?
\( x \) | 2.9 | 2.95 | 2.99 | 2.999 | 3.001 | 3.01 | 3.05 | 3.1 |
---|---|---|---|---|---|---|---|---|
\( f(x) \) | 3.4 | 3.1 | 3.004 | 3.00004 | 3.00004 | 3.004 | 3.1 | 3.4 |
A .
B
C
D
▶️ Answer/Explanation
Solution
1. Table Analysis:
- As \( x \) approaches 3 from both sides, \( f(x) \) approaches 3.
- The values are symmetric around \( x = 3 \), suggesting continuity.
2. Graph Requirements:
- The correct graph must show \( f(x) \rightarrow 3 \) as \( x \rightarrow 3 \).
- Graph D shows a jump discontinuity at \( x = 3 \), which contradicts the table data.
3. Conclusion:
Graph D could not represent \( f \) because it violates the limit behavior shown in the table.
✅ Answer: D)