AP Calculus AB

Limits and Continuity: Laws, Definitions, and Asymptotes

Calculus AB

  • ✓By the end of this lesson students will be able to apply limit laws to evaluate limits of functions.
  • ✓By the end of this lesson students will be able to determine the continuity of a function at a point and on an interval using the formal definition of continuity.
  • ✓By the end of this lesson students will be able to identify and describe vertical and horizontal asymptotes of functions.
  • ✓By the end of this lesson students will be able to analyze the behavior of functions as x approaches infinity or negative infinity.

Key concepts

Limit Laws

Limit laws allow us to evaluate limits of complex functions by breaking them down into simpler parts. These laws apply when the individual limits exist.

Let c be a constant and assume that lim x→a f(x) and lim x→a g(x) exist.\n1. Sum Rule: lim x→a [f(x) + g(x)] = lim x→a f(x) + lim x→a g(x)\n2. Difference Rule: lim x→a [f(x) - g(x)] = lim x→a f(x) - lim x→a g(x)\n3. Constant Multiple Rule: lim x→a [c * f(x)] = c * lim x→a f(x)\n4. Product Rule: lim x→a [f(x) * g(x)] = lim x→a f(x) * lim x→a g(x)\n5. Quotient Rule: lim x→a [f(x) / g(x)] = [lim x→a f(x)] / [lim x→a g(x)], provided lim x→a g(x) ≠ 0\n6. Power Rule: lim x→a [f(x)]^n = [lim x→a f(x)]^n, for any positive integer n\n7. Root Rule: lim x→a [n√f(x)] = n√[lim x→a f(x)], for any positive integer n (if n is even, assume lim x→a f(x) > 0)
Continuity at a Point

A function f(x) is continuous at a point x = c if and only if all three of the following conditions are met. If any condition fails, the function is discontinuous at c.

1. f(c) is defined.\n2. lim x→c f(x) exists.\n3. lim x→c f(x) = f(c).
Types of Discontinuities

Discontinuities are points where a function is not continuous. There are three main types:\n1. Removable Discontinuity (Hole): Occurs when lim x→c f(x) exists but f(c) is undefined or lim x→c f(x) ≠ f(c). This often happens when a common factor can be canceled from the numerator and denominator of a rational function.\n2. Jump Discontinuity: Occurs when the left-hand limit and the right-hand limit at x=c both exist but are not equal (lim x→c- f(x) ≠ lim x→c+ f(x)). This is common in piecewise functions.\n3. Infinite Discontinuity: Occurs when lim x→c f(x) = ±∞. This is associated with vertical asymptotes.

Vertical Asymptotes

A vertical line x=c is a vertical asymptote of the graph of a function f(x) if f(x) approaches positive or negative infinity as x approaches c from either the left or the right. For rational functions, vertical asymptotes occur at values of x where the denominator is zero and the numerator is non-zero.

lim x→c+ f(x) = ±∞ OR lim x→c- f(x) = ±∞
Horizontal Asymptotes

A horizontal line y=L is a horizontal asymptote of the graph of a function f(x) if f(x) approaches L as x approaches positive or negative infinity. Horizontal asymptotes describe the end behavior of a function. For rational functions P(x)/Q(x):\n1. If degree(P) < degree(Q), then y=0 is the horizontal asymptote.\n2. If degree(P) = degree(Q), then y = (leading coefficient of P) / (leading coefficient of Q) is the horizontal asymptote.\n3. If degree(P) > degree(Q), there is no horizontal asymptote (though there might be a slant asymptote).

lim x→∞ f(x) = L OR lim x→-∞ f(x) = L

Key facts to remember

  • 1Limit laws simplify the evaluation of limits by allowing operations on individual limits.
  • 2A function is continuous at a point if the function is defined, the limit exists, and the limit equals the function value at that point.
  • 3Vertical asymptotes occur where a function's value approaches ±∞, often when the denominator of a rational function is zero and the numerator is non-zero.
  • 4Horizontal asymptotes describe the end behavior of a function as x approaches ±∞, indicating a value the function approaches.
  • 5Removable discontinuities are 'holes' in the graph where the limit exists but does not match the function value or the function is undefined.
  • 6Jump discontinuities occur when the left-hand and right-hand limits at a point are different.
  • 7Infinite discontinuities are associated with vertical asymptotes.

Worked examples

Example 1

Evaluate the limit: lim x→2 (3x^2 - 5x + 1) / (x - 4)

IFirst, attempt direct substitution to see if the limit can be found directly and to check if the denominator is zero.
IISubstitute x=2 into the numerator: 3(2)^2 - 5(2) + 1 = 3(4) - 10 + 1 = 12 - 10 + 1 = 3.
IIISubstitute x=2 into the denominator: 2 - 4 = -2.
IVSince the limit of the denominator is not zero (-2 ≠ 0), we can apply the Quotient Rule for limits.
Vlim x→2 (3x^2 - 5x + 1) / (x - 4) = [lim x→2 (3x^2 - 5x + 1)] / [lim x→2 (x - 4)]
VIUsing the Sum, Difference, Constant Multiple, and Power Rules: = 3 / (-2)

Answer

-3/2 or -1.5

Direct substitution is the first method to try for limits of polynomials and rational functions. If the denominator is non-zero, the limit is simply the value of the function at that point.

Example 2

Determine if the function f(x) = { (x^2 - 9) / (x - 3) for x ≠ 3 ; 6 for x = 3 } is continuous at x = 3. If not, identify the type of discontinuity.

ICheck Condition 1: Is f(3) defined?
IIFrom the definition, f(3) = 6. So, f(3) is defined.
IIICheck Condition 2: Does lim x→3 f(x) exist?
IVFor x ≠ 3, f(x) = (x^2 - 9) / (x - 3). We can simplify this expression: (x^2 - 9) / (x - 3) = (x - 3)(x + 3) / (x - 3) = x + 3 (for x ≠ 3).
VNow, evaluate the limit: lim x→3 (x + 3) = 3 + 3 = 6. So, lim x→3 f(x) exists and equals 6.
VICheck Condition 3: Is lim x→3 f(x) = f(3)?
VIIWe found lim x→3 f(x) = 6 and f(3) = 6. Since 6 = 6, this condition is met.
VIIISince all three conditions for continuity are met, the function is continuous at x = 3.

Answer

The function f(x) is continuous at x = 3.

If f(3) had been a different value (e.g., 5), then Condition 3 would fail, and it would be a removable discontinuity. If f(3) were undefined, Condition 1 would fail, also a removable discontinuity.

Example 3

Find all vertical and horizontal asymptotes of the function g(x) = (2x^2 + 5x - 3) / (x^2 - 4).

ITo find Vertical Asymptotes (VAs), set the denominator equal to zero and solve for x. Then, check if the numerator is non-zero at these x-values.
IIDenominator: x^2 - 4 = 0 => (x - 2)(x + 2) = 0 => x = 2, x = -2.
IIICheck numerator at x = 2: 2(2)^2 + 5(2) - 3 = 2(4) + 10 - 3 = 8 + 10 - 3 = 15. Since 15 ≠ 0, x = 2 is a VA.
IVCheck numerator at x = -2: 2(-2)^2 + 5(-2) - 3 = 2(4) - 10 - 3 = 8 - 10 - 3 = -5. Since -5 ≠ 0, x = -2 is a VA.
VTo find Horizontal Asymptotes (HAs), evaluate the limit of g(x) as x approaches ±∞.
VICompare the degrees of the numerator and denominator. The degree of the numerator (2x^2 + 5x - 3) is 2. The degree of the denominator (x^2 - 4) is 2.
VIISince the degrees are equal, the HA is y = (leading coefficient of numerator) / (leading coefficient of denominator).
VIIILeading coefficient of numerator = 2. Leading coefficient of denominator = 1.
9So, the HA is y = 2/1 = 2.

Answer

Vertical asymptotes at x = 2 and x = -2. Horizontal asymptote at y = 2.

Always check if any factors cancel between the numerator and denominator before identifying vertical asymptotes. If a factor cancels, it indicates a hole, not a vertical asymptote.

Common mistakes

  • ✗Incorrectly applying the Quotient Rule for limits when the limit of the denominator is zero, leading to an undefined result rather than an indeterminate form requiring further analysis.
  • ✗Forgetting to check all three conditions for continuity at a point, especially the condition that the limit must equal the function value.
  • ✗Confusing a removable discontinuity (a hole in the graph) with a vertical asymptote. A hole occurs when a factor cancels out, while a vertical asymptote occurs when a factor remains in the denominator.
  • ✗Incorrectly determining horizontal asymptotes, particularly for rational functions where the degrees of the numerator and denominator are not properly compared.
  • ✗Not using proper limit notation (e.g., writing 'f(x) = L' instead of 'lim x→a f(x) = L') throughout the solution process, which can lead to loss of points on exams.

Exam tips

  • ★Always use correct limit notation (e.g., lim x→c f(x)) in every step of your solution when evaluating limits.
  • ★For continuity questions, explicitly state and verify each of the three conditions for continuity at the specified point.
  • ★When evaluating limits, always try direct substitution first. If it results in an indeterminate form (e.g., 0/0), then consider algebraic manipulation (factoring, rationalizing, finding a common denominator).
  • ★To find vertical asymptotes, factor the numerator and denominator completely. Any factor remaining in the denominator after cancellation corresponds to a vertical asymptote. Any canceled factor corresponds to a hole.
  • ★To find horizontal asymptotes, evaluate the limit as x approaches positive and negative infinity. Remember to consider the degrees of the polynomials for rational functions.

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