Calculus 1 Quiz: Estimating Limits From Graphs
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Estimating Limits From GraphsQuestion 1 of 19

The graph of f(x) is such that lim(x→-1) f(x) = 8. A new function g(x) is created whose graph is that of f(x) shifted 4 units to the left and 2 units down. What is lim(x→-5) g(x)?

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Calculus 1 Quiz

Calculus 1 Quiz: Estimating Limits From Graphs

Practice Estimating Limits From Graphs in Calculus 1 with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Estimating Limits From Graphs, giving you a quick way to practice the rules, question types, and explanations that matter most for Calculus 1.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

All questions

Question 1

The graph of f(x) is such that lim(x→-1) f(x) = 8. A new function g(x) is created whose graph is that of f(x) shifted 4 units to the left and 2 units down. What is lim(x→-5) g(x)?

  1. 4
  2. 6 (correct answer)
  3. 8
  4. 10
Explanation: The transformation described means that g(x) = f(x+4) - 2. We want to find lim(x→-5) g(x) = lim(x→-5) [f(x+4) - 2]. Let u = x+4. As x → -5, u → -5+4 = -1. The limit becomes lim(u→-1) [f(u) - 2] = [lim(u→-1) f(u)] - 2. We are given that lim(x→-1) f(x) = 8, so lim(u→-1) f(u) = 8. The final result is 8 - 2 = 6.

Question 2

The graph of g(x) shows that as x approaches 2, g(x) approaches 3, but always from values less than 3. The graph of f(u) is defined by a jump discontinuity at u = 3; for u < 3, f(u) = u + 2, and for u ≥ 3, f(u) = 2u - 3. What is lim(x→2) f(g(x))?

  1. 3
  2. 5 (correct answer)
  3. Does not exist
  4. Cannot be determined
Explanation: Let u = g(x). The stem states that as x approaches 2, g(x) approaches 3 from below (g(x) < 3). This means u approaches 3 from the left side (u → 3⁻). Therefore, we need to evaluate lim(u→3⁻) f(u). According to the definition of f(u), for u < 3, we use the expression f(u) = u + 2. So, lim(u→3⁻) f(u) = 3 + 2 = 5. A common error is to evaluate the right-hand limit, which would give 2(3) - 3 = 3.

Question 3

The graph of f(x) is continuous for all x. The graph of g(x) is identical to the graph of f(x) for all x except at x = a, where the graph of g(x) has a removable discontinuity. Which of the following must be equal to f(a)?

  1. g(a)
  2. lim(x→a⁻) g(x) (correct answer)
  3. lim(x→∞) g(x)
  4. The slope of g(x) at x=a
Explanation: Since g(x) is identical to the continuous function f(x) everywhere except at x=a, the limit of g(x) as x approaches 'a' must be equal to the value of f(x) at 'a'. That is, lim(x→a) g(x) = f(a). Since the limit exists (because it's a removable discontinuity), the left-hand limit equals the right-hand limit, and both are equal to the overall limit. Therefore, lim(x→a⁻) g(x) = f(a). Choice A is incorrect because at a removable discontinuity, g(a) is either undefined or not equal to the limit. Choices C and D are irrelevant to the value of f at a.

Question 4

The graph of f(x) has a vertical asymptote at x = 0, where lim(x→0) f(x) = ∞. The graph of g(x) is continuous and passes through the origin, so lim(x→0) g(x) = 0. Based on this information, what is lim(x→0) [g(x) / f(x)]?

  1. 0 (correct answer)
  2. 1
  3. ∞
  4. Cannot be determined
Explanation: We are evaluating the limit of a quotient. The limit of the numerator is lim(x→0) g(x) = 0. The limit of the denominator is lim(x→0) f(x) = ∞. A limit of the form 0/∞ evaluates to 0. As the numerator approaches zero and the denominator grows infinitely large, the fraction's value approaches zero.

Question 5

The graph of a function h(x) oscillates with decreasing amplitude around the horizontal line y = 5. As x approaches 2 from either side, the oscillations become infinitesimally small, converging on the point (2, 5). What is lim(x→2) h(x)?

  1. 0
  2. 2
  3. Does not exist
  4. 5 (correct answer)
Explanation: When you encounter a limit problem involving oscillating functions, focus on where the function values are heading, not the path they take to get there. The key insight is that limits depend only on the behavior of function values as you approach the target point, regardless of how wildly the function might oscillate along the way. The problem tells you that as x approaches 2 from either side, the oscillations become "infinitesimally small" and converge on the point (2, 5). This means that no matter how much h(x) oscillates, those oscillations are getting smaller and smaller, and the function values are getting arbitrarily close to y = 5. Since this happens from both the left and right sides of x = 2, we have lim⁡x→2h(x)=5\lim_{x \to 2} h(x) = 5. Answer (A) suggests 0, which might tempt you if you're thinking about the amplitude of oscillations approaching zero, but that's not what the limit measures. Answer (B) suggests 2, which confuses the x-coordinate we're approaching with the y-value the function approaches. Answer (C) claims the limit doesn't exist, which would only be true if the function approached different values from the left and right sides, or if the oscillations weren't diminishing. The correct answer is (D) 5, because both one-sided limits equal 5. Study tip: For oscillating functions, always ask yourself: "What value are the oscillations centered around, and are they getting smaller?" If they're shrinking toward a specific horizontal line, that's your limit.

Question 6

The graph of a function f(x) is described as follows: For x < 2, the graph follows the parabola y = x². For x > 2, the graph follows the line y = x + 2. At x = 2, the function's value is given by a point at (2, 1). Based on this graphical description, what is the value of lim(x→2) f(x)?

  1. 1
  2. 2
  3. 4 (correct answer)
  4. Does not exist
Explanation: To find the limit as x approaches 2, we must examine the left-hand and right-hand limits. The left-hand limit is determined by the parabola y = x²: lim(x→2⁻) f(x) = 2² = 4. The right-hand limit is determined by the line y = x + 2: lim(x→2⁺) f(x) = 2 + 2 = 4. Since the left-hand and right-hand limits are equal, the limit exists and is 4. The function's value at x=2, which is f(2)=1, does not affect the value of the limit.

Question 7

The graph of f(x) is a continuous curve that passes through the point (3, 5). The graph of g(x) follows the line y = 2x - 4 for all x ≠ 3, but has a removable discontinuity (a hole) at x = 3, and g(3) = 1. What is the value of lim(x→3) [f(x) * g(x)]?

  1. 2
  2. 5
  3. 10 (correct answer)
  4. Does not exist
Explanation: Using the limit properties, lim(x→3) [f(x) * g(x)] = [lim(x→3) f(x)] * [lim(x→3) g(x)]. Since f(x) is continuous, lim(x→3) f(x) = f(3) = 5. For g(x), the limit is the value approached by the line at x = 3, not the function value g(3). So, lim(x→3) g(x) = 2(3) - 4 = 2. Therefore, the limit of the product is 5 * 2 = 10. A common mistake is to use g(3)=1, which would yield 5 * 1 = 5.

Question 8

The graph of a function f(x) consists of a line segment from (-4, 5) to an open circle at (-1, 2), and another line segment starting from a closed circle at (-1, 4) to (2, 1). What is the value of lim(x→-1⁻) f(x) + lim(x→-1⁺) f(x) + f(-1)?

  1. 8
  2. 9
  3. 10 (correct answer)
  4. 11
Explanation: We need to evaluate three components from the graph's description. The left-hand limit, lim(x→-1⁻) f(x), is the value approached by the first segment, which ends at the open circle at (-1, 2). So, the limit is 2. The right-hand limit, lim(x→-1⁺) f(x), is the value approached from the start of the second segment, the closed circle at (-1, 4). So, the limit is 4. The function value, f(-1), is the y-coordinate of the closed circle, which is 4. The sum is 2 + 4 + 4 = 10.

Question 9

The graph of the function h(x) has a horizontal asymptote at y = -2 as x → ∞. The graph of the function k(x) has a horizontal asymptote at y = 3 as x → ∞. Based on this information, what is lim(x→∞) [h(x) + 2k(x)]?

  1. -2
  2. 1
  3. 4 (correct answer)
  4. Cannot be determined
Explanation: The existence of horizontal asymptotes gives us the limits at infinity. We have lim(x→∞) h(x) = -2 and lim(x→∞) k(x) = 3. Using the properties of limits, we can find the combined limit: lim(x→∞) [h(x) + 2k(x)] = lim(x→∞) h(x) + 2 * lim(x→∞) k(x) = -2 + 2 * 3 = -2 + 6 = 4.

Question 10

The graph of a function k(x) consists of two line segments meeting at a sharp corner at (-1, 3). For x < -1, the slope of the line is 2. For x > -1, the slope is -1. What is the value of lim(x→-1) k(x)?

  1. -1
  2. 2
  3. 3 (correct answer)
  4. Does not exist
Explanation: The limit of a function at a point is the y-value that the function approaches from both the left and the right. The description states that the two line segments meet at the point (-1, 3). This means that as x approaches -1 from the left and from the right, the graph of the function approaches the y-value of 3. Therefore, lim(x→-1) k(x) = 3. The sharp corner indicates that the function is not differentiable at x = -1, but it is continuous, and the limit exists.

Question 11

The graph of g(x) approaches the value 4 as x approaches 1. The graph of f(u) is described by the equation y = u² for u ≤ 4 and y = 2u + 8 for u > 4. What is lim(x→1) f(g(x))?

  1. 16 (correct answer)
  2. 32
  3. Does not exist
  4. Cannot be determined without knowing if g(x) approaches 4 from above or below
Explanation: Let u = g(x). As x → 1, u → 4. We need to find lim(u→4) f(u). Let's check the one-sided limits for f(u). lim(u→4⁻) f(u) is governed by u², so the limit is 4² = 16. lim(u→4⁺) f(u) is governed by 2u + 8, so the limit is 2(4) + 8 = 16. Since both one-sided limits of f(u) at u=4 are 16, lim(u→4) f(u) = 16. Thus, lim(x→1) f(g(x)) = 16. It is not necessary to know the direction of approach for g(x) because f(u) is continuous at u=4.

Question 12

The graph of a function g(x) has a vertical asymptote at x=2. As x approaches 2 from either the left or the right, the graph of g(x) increases without bound. What is lim(x→2) [1 / g(x)]?

  1. 0 (correct answer)
  2. 1
  3. 2
  4. ∞
Explanation: The graphical description tells us that lim(x→2⁻) g(x) = ∞ and lim(x→2⁺) g(x) = ∞. Therefore, the two-sided limit is lim(x→2) g(x) = ∞. We are asked to find the limit of 1 / g(x). As the denominator g(x) grows infinitely large, the value of the fraction 1 / g(x) approaches 0. Thus, lim(x→2) [1 / g(x)] = 0.

Question 13

The graph of f(x) shows that lim(x→∞) f(x) = 3. The graph of g(x) shows that lim(x→1⁺) g(x) = ∞. What can be concluded about lim(x→1⁺) f(g(x))?

  1. 1
  2. 3 (correct answer)
  3. ∞
  4. Cannot be determined
Explanation: This is a limit of a composite function. Let u = g(x). As x approaches 1 from the right (x → 1⁺), the value of u = g(x) approaches ∞. Therefore, the problem transforms into finding the limit of f(u) as u approaches ∞. We are given from the graph of f(x) that lim(x→∞) f(x) = 3, which means lim(u→∞) f(u) = 3. So, lim(x→1⁺) f(g(x)) = 3.

Question 14

The graph of f(x) is given by the line y = x - 2. What is the value of lim(x→2⁻) (x² - 4) / f(x)?

  1. -4
  2. 0
  3. Does not exist
  4. 4 (correct answer)
Explanation: This is a limit problem involving a rational function where you need to evaluate the behavior as x approaches 2 from the left. Since f(x) = x - 2, you're looking at lim⁡x→2−x2−4x−2\lim_{x \to 2^-} \frac{x^2 - 4}{x - 2}. The key insight is recognizing that the numerator x² - 4 can be factored as a difference of squares: x² - 4 = (x + 2)(x - 2). This gives you lim⁡x→2−(x+2)(x−2)x−2\lim_{x \to 2^-} \frac{(x + 2)(x - 2)}{x - 2}. Since you're approaching 2 but not actually at x = 2, you can cancel the common factor (x - 2), leaving lim⁡x→2−(x+2)\lim_{x \to 2^-} (x + 2). Now the limit is straightforward: as x approaches 2, (x + 2) approaches 4. Looking at the wrong answers: Choice A (-4) likely comes from incorrectly substituting x = 2 directly into the original expression without factoring, getting 0/0 and then making a sign error. Choice B (0) represents the trap of thinking that since both numerator and denominator approach 0, the limit must be 0 - this ignores the cancellation that occurs after factoring. Choice C (does not exist) might result from incorrectly believing that the 0/0 indeterminate form means no limit exists, rather than requiring further analysis through algebraic manipulation. Strategy tip: When you encounter 0/0 indeterminate forms, always try factoring first. Most calculus problems are designed so that common factors will cancel, revealing the true limit value.

Question 15

The graph of f(x) is a line y = x - 4. Consider the function g(x) = f(x)/|f(x)|. From a graphical perspective, what is lim(x→4⁻) g(x)?

  1. -4
  2. -1 (correct answer)
  3. 1
  4. Does not exist
Explanation: The function is g(x) = (x-4)/|x-4|. We are evaluating the limit as x approaches 4 from the left (x → 4⁻). For values of x less than 4, the term (x-4) is negative. Therefore, by the definition of absolute value, |x-4| = -(x-4) for x < 4. Substituting this into the expression for g(x), we get g(x) = (x-4)/(-(x-4)) = -1 for all x < 4. Thus, the limit as x approaches 4 from the left is -1.

Question 16

The graph of f(x) is such that lim(x→3) f(x) = 6 and f(3) = 6. What can be concluded about the graph of f(x) at x=3?

  1. The graph has a removable discontinuity at x=3.
  2. The graph has a jump discontinuity at x=3.
  3. The graph has a vertical asymptote at x=3.
  4. The graph is continuous at x=3. (correct answer)
Explanation: When you encounter limit and function value problems, you're being tested on the precise definition of continuity. A function is continuous at a point when three conditions are met: the function is defined at that point, the limit exists, and the limit equals the function value. Let's examine what we know: lim⁡x→3f(x)=6\lim_{x \to 3} f(x) = 6 and f(3)=6f(3) = 6. Since the limit as x approaches 3 equals 6, and the function value at x = 3 also equals 6, all three continuity conditions are satisfied. The function is defined at x = 3, the limit exists, and lim⁡x→3f(x)=f(3)=6\lim_{x \to 3} f(x) = f(3) = 6. This means the graph is continuous at x = 3, making D correct. Now let's see why the other options are wrong. Option A describes a removable discontinuity, which occurs when the limit exists but either the function isn't defined at the point or the function value differs from the limit. Since f(3)=lim⁡x→3f(x)=6f(3) = \lim_{x \to 3} f(x) = 6, there's no discontinuity to remove. Option B, a jump discontinuity, happens when left and right limits exist but are unequal. Since our limit exists and equals 6, there's no jump. Option C suggests a vertical asymptote, which occurs when limits approach infinity. Our limit approaches the finite value 6, not infinity. Remember this key pattern: if the limit exists and equals the function value at that point, you have continuity. Watch for problems that give you these two pieces of information—they're testing whether you recognize the continuity definition.

Question 17

The graph of f(x) is squeezed between the graphs of g(x) = -x² + 6x - 7 and h(x) = x² - 6x + 11 in a neighborhood around x = 3. What is lim(x→3) f(x)?

  1. 1
  2. 2 (correct answer)
  3. 3
  4. Cannot be determined
Explanation: This question describes the conditions for the Squeeze Theorem. We must find the limits of the bounding functions as x approaches 3. lim(x→3) g(x) = -(3)² + 6(3) - 7 = -9 + 18 - 7 = 2. lim(x→3) h(x) = (3)² - 6(3) + 11 = 9 - 18 + 11 = 2. Since f(x) is between g(x) and h(x), and both g(x) and h(x) approach 2 as x approaches 3, the Squeeze Theorem implies that lim(x→3) f(x) must also be 2.

Question 18

The graph of a function f(x) has a horizontal asymptote y = 4 as x → -∞, and it decreases without bound as x → ∞. What is lim(x→∞) f(x) - lim(x→-∞) f(x)?

  1. 0
  2. -4
  3. ∞
  4. -∞ (correct answer)
Explanation: From the graphical description, we extract the limits. The horizontal asymptote at y=4 as x → -∞ means lim(x→-∞) f(x) = 4. The statement that the graph decreases without bound as x → ∞ means lim(x→∞) f(x) = -∞. The expression to evaluate is lim(x→∞) f(x) - lim(x→-∞) f(x) = (-∞) - 4. Subtracting a finite number from negative infinity still results in negative infinity. Thus, the answer is -∞.

Question 19

The graph of f(x) is a semicircle defined by y = √(4 - x²) for -2 ≤ x ≤ 2. What is lim(x→-2⁺) f(x)?

  1. -2
  2. 0 (correct answer)
  3. 2
  4. Does not exist
Explanation: The function f(x) = √(4 - x²) describes the upper half of a circle with radius 2 centered at the origin. The domain is [-2, 2]. The question asks for the right-hand limit at x = -2. As x approaches -2 from the right side (from values like -1.9, -1.99, etc.), we can substitute the value into the continuous function: lim(x→-2⁺) √(4 - x²) = √(4 - (-2)²) = √(4 - 4) = 0. The limit exists because we are only considering the approach from within the function's domain.