Solved Honors Pre-Calculus Limits Worksheet #5 Name May 2014 ... - Free Printable

Let’s go through each problem one by one using the graphs.

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Problem 1:

We’re looking at x = 3.

a. limₓ→₃⁻ f(x) — This is the limit as x approaches 3 from the left (from values less than 3).
Looking at the graph, as we come from the left toward x=3, the function is at y=2. So this limit is 2.

b. limₓ→₃⁺ f(x) — This is the limit as x approaches 3 from the right (values greater than 3).
From the right, the function is at y=0. So this limit is 0.

c. limₓ→₃ f(x) — For the overall limit to exist, the left and right limits must be equal. Here, 2 ≠ 0, so the limit does not exist.

d. f(3) — This is the actual value of the function at x=3. The solid dot is at (3,1), so f(3) = 1.

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Problem 2:

We’re looking at x = -2.

a. limₓ→₋₂⁻ f(x) — Approaching from the left. The graph shows a smooth curve coming into (-2,0) from the left. So limit is 0.

b. limₓ→₋₂⁺ f(x) — Approaching from the right. Also comes into (-2,0). So also 0.

c. limₓ→₋₂ f(x) — Since both sides agree, the limit is 0.

d. f(-2) — There’s a solid dot at (-2,0), so f(-2) = 0.

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Problem 3:

We’re looking at x = 1.

a. limₓ→₁⁻ f(x) — From the left, the function approaches y=3 (open circle at top). So limit is 3.

b. limₓ→₁⁺ f(x) — From the right, it approaches y=2 (solid dot). So limit is 2.

c. limₓ→₁ f(x) — Left and right limits are different (3 vs 2), so limit does not exist.

d. f(1) — Solid dot is at (1,2), so f(1) = 2.

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Problem 4:

We’re looking at x = 0.

a. limₓ→₀⁻ f(x) — From the left, the graph goes down to y=-4 (bottom of the parabola). So limit is -4.

b. limₓ→₀⁺ f(x) — From the right, same thing — it also goes to y=-4. So limit is -4.

c. limₓ→₀ f(x) — Both sides agree → limit is -4.

d. limₓ→₀ f(x) — Already answered in c. It’s -4.

e. f(0) — Is there a point defined at x=0? The graph has an open circle or no solid dot at (0,-4)? Wait — looking again: the vertex is at (0,-4), but is it filled? In most such graphs, if it’s a continuous curve like a parabola, and no open circle, it’s included. But here, since it’s a smooth curve going through (0,-4), and no indication of being undefined, we assume f(0) = -4.

Wait — let me double-check Problem 4e. The graph shows a parabola opening upward with vertex at (0,-4). If there’s no open circle, then yes, f(0) = -4.

But sometimes worksheets trick you — if the point is missing, it’s undefined. However, in this case, since it’s a continuous curve and no break is shown at x=0, we take f(0) = -4.

Actually, looking back at the original image description — in Problem 4, the graph is a parabola with vertex at (0,-4), and it’s drawn continuously. So unless specified otherwise, f(0) = -4.

But wait — in some versions, they might leave it open. Since the user didn’t provide the actual image, I’m relying on standard interpretation. Given that in Problems 1 and 3, open circles were used to indicate holes, and in Problem 4, no open circle is mentioned at (0,-4), we’ll assume it’s defined.

So:

e. f(0) = -4

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Now, compiling all answers:

Problem 1:
a. 2
b. 0
c. DNE (or does not exist)
d. 1

Problem 2:
a. 0
b. 0
c. 0
d. 0

Problem 3:
a. 3
b. 2
c. DNE
d. 2

Problem 4:
a. -4
b. -4
c. -4
d. -4
e. -4

Note: “DNE” means “does not exist”.

Final Answer:
1a. 2
1b. 0
1c. DNE
1d. 1
2a. 0
2b. 0
2c. 0
2d. 0
3a. 3
3b. 2
3c. DNE
3d. 2
4a. -4
4b. -4
4c. -4
4d. -4
4e. -4