Source: Textbook Section 1.5, Recitation Review 1
Tags: limits, one-sided limits, two-sided limits, vertical asymptotes, infinite limits, rational functions, Calculus I, MTH 132, MSU
Difficulty: Introductory to Intermediate Prerequisites: Section 1.4 (average rate of change), basic algebra, familiarity with rational functions and piecewise functions.
A limit describes what value a function approaches as x gets close to some number, even if the function never reaches that value. One-sided limits look at approach from the left or right only. Vertical asymptotes occur where a function blows up to positive or negative infinity, typically where a denominator equals zero and the numerator does not.
Limit (two-sided)
We write lim(x->c) f(x) = L to mean that f(x) gets arbitrarily close to L as x approaches c from both sides. Think of it as: what y-value is the function heading towards as x sneaks up on c?
Left-hand limit (one-sided)
Written lim(x->c-) f(x). This considers only values of x that are less than c (approaching from the left on the number line). In simple terms, this means: cover the right side of the graph and look only at what is happening to the left of c.
Right-hand limit (one-sided)
Written lim(x->c+) f(x). This considers only values of x that are greater than c (approaching from the right). In simple terms, this means: cover the left side of the graph and look only at what is happening to the right of c.
Two-sided limit exists
The two-sided limit lim(x->c) f(x) = L exists if and only if both one-sided limits exist and are equal: lim(x->c-) f(x) = lim(x->c+) f(x) = L. Think of it as: the function has to be heading to the same place from both directions.
Vertical asymptote
A vertical line x = a where f(x) approaches positive infinity or negative infinity (or both) as x approaches a from one or both sides. In simple terms, this means: the graph shoots up or down without bound near that x-value, and the function is never defined there.
Infinite limit
When we write lim(x->c) f(x) = +infinity or -infinity, we mean the function grows without bound. The limit does not exist as a finite number, but we use the infinity notation to describe the behaviour.
To find lim(x->c) f(x) from a graph, trace the curve from both sides toward x = c and see what y-value the curve approaches.
An open circle at a point means the function is not defined there (or is defined at a different value). The limit still exists if the curve approaches a single y-value from both sides.
A filled circle shows the actual function value f(c). This value does not affect the limit unless it happens to match what the curve approaches.
The left-hand limit lim(x->c-) f(x) looks at what happens as x approaches c through values smaller than c.
The right-hand limit lim(x->c+) f(x) looks at what happens as x approaches c through values larger than c.
The two-sided limit exists only when both one-sided limits are equal. If they disagree, the two-sided limit does not exist (DNE).
For a piecewise function like g(x) = { -x for x < 3, -1 for x = 3, x^2 - 13 for x > 3 }:
Left-hand limit as x->3: use the rule for x < 3, so lim(x->3-) g(x) = -3.
Right-hand limit as x->3: use the rule for x > 3, so lim(x->3+) g(x) = 9 - 13 = -4.
Since -3 and -4 are not equal, lim(x->3) g(x) does not exist.
The value g(3) = -1 is irrelevant to the limit. Limits care about what happens near the point, not at it.
A rational function p(x)/q(x) has a vertical asymptote at x = a if q(a) = 0 and p(a) is not 0 (after cancelling any common factors).
Step-by-step method:
Factor both numerator and denominator completely.
Cancel any common factors.
Set the remaining denominator equal to zero. Those x-values are the vertical asymptotes.
If a factor cancels from both top and bottom, that x-value is a hole (removable discontinuity), not a vertical asymptote.
f(x) = 1 / [(x - 1)(x + 3)]
No common factors. Denominator is zero when x = 1 and x = -3. Vertical asymptotes at x = 1 and x = -3.
g(x) = (x - 1) / [(x - 1)(x + 3)]
The factor (x - 1) cancels. After cancellation: g(x) = 1/(x + 3). Vertical asymptote only at x = -3. There is a hole at x = 1.
h(x) = (x^2 + 3x + 2) / (x^2 - 4)
Factor: (x + 1)(x + 2) / [(x - 2)(x + 2)]. Cancel (x + 2). After cancellation: (x + 1)/(x - 2). Vertical asymptote at x = 2. Hole at x = -2.
For f(x) = 1/x^2:
lim(x->0-) 1/x^2 = +infinity (x^2 is always positive, so the function goes up on both sides)
lim(x->0+) 1/x^2 = +infinity
lim(x->0) 1/x^2 = +infinity (both sides agree)
Vertical asymptote at x = 0.
For f(x) = 1/(x - 5):
lim(x->5-) 1/(x - 5) = -infinity (denominator is small and negative)
lim(x->5+) 1/(x - 5) = +infinity (denominator is small and positive)
lim(x->5) 1/(x - 5) does not exist (the two sides disagree in sign)
Vertical asymptote at x = 5.
For f(x) = x^3 sec(x) on [0, pi]:
sec(x) = 1/cos(x), so vertical asymptotes occur where cos(x) = 0. On [0, pi], that is x = pi/2.
lim(x->(pi/2)-) x^3 sec(x) = +infinity (cos(x) is positive and approaching 0 from the left of pi/2)
lim(x->(pi/2)+) x^3 sec(x) = -infinity (cos(x) is negative just past pi/2)
\lim_{x \to c} f(x) = L \quad \text{exists} \iff \quad \lim_{x \to c^-} f(x) = \lim_{x \to c^+} f(x) = LVertical asymptote test for p(x)/q(x): factor, cancel common factors, then set the remaining denominator to zero.
Sign analysis shortcut for infinite limits: check whether the denominator is positive or negative just to the left and just to the right of the asymptote. Positive denominator near zero gives +infinity; negative gives -infinity.
Vertical asymptotes model real situations where a quantity grows without bound as a variable approaches some threshold. In physics, the electric field strength near a point charge follows an inverse-square law and tends to infinity as you approach the charge. In economics, certain cost functions blow up as a resource approaches full depletion. Recognising where a model breaks down (goes to infinity) is a practical skill, not just an exam topic.
Students often think the limit of f at x = c must equal f(c). It does not. The limit is about the behaviour near c, not at c. A function can have a limit of 5 at x = 2 even if f(2) = 100 or f(2) is undefined.
Students confuse "the limit is infinity" with "the limit exists." When we write lim = +infinity, we are describing unbounded behaviour. Technically, the limit does not exist as a real number.
When finding vertical asymptotes, students sometimes forget to cancel common factors first. If both the numerator and denominator are zero at the same point, you have a hole, not an asymptote. Always factor and simplify before declaring asymptotes.
Students sometimes check only one side of a potential asymptote. You need to check both sides to fully describe the behaviour: the function can go to +infinity on one side and -infinity on the other.
⚠️ Expect a graph where you must read off several limits (one-sided and two-sided) at different x-values. Practise distinguishing open vs filled circles.
⚠️ Piecewise limit problems are very common. You will be given a function with two or three pieces and asked whether the limit exists at the boundary. Always check both sides separately.
⚠️ Finding vertical asymptotes of rational functions appears on nearly every exam. The procedure (factor, cancel, set remaining denominator to zero) must be automatic.
⚠️ Know how to determine the sign of an infinite limit (positive vs negative infinity) by checking the sign of the denominator on each side of the asymptote.
True or False: If lim(x->c-) f(x) = 3 and lim(x->c+) f(x) = 5, then lim(x->c) f(x) = 4.
True or False: A vertical asymptote occurs at every x-value where the denominator of a rational function is zero.
Fill in the blank: If f(x) = 1/(x - 5), then lim(x->5+) f(x) = ___.
True or False: The value f(c) determines the limit of f at c.
Fill in the blank: To find vertical asymptotes, first ___ the numerator and denominator.
Answers: 1. False (the two-sided limit does not exist because the one-sided limits disagree). 2. False (you must cancel common factors first; a common factor gives a hole). 3. +infinity. 4. False. 5. Factor.
Q: For the piecewise function g(x) = { -x if x < 3, -1 if x = 3, x^2 - 13 if x > 3 }, does lim(x->3) g(x) exist? What is g(3)?
A: lim(x->3-) g(x) = -3 and lim(x->3+) g(x) = -4. Since the one-sided limits are not equal, the two-sided limit does not exist. g(3) = -1, but that value does not affect the limit.
Q: Find the vertical asymptotes of h(x) = (x^2 + 3x + 2)/(x^2 - 4).
A: Factor: (x+1)(x+2)/[(x-2)(x+2)]. Cancel (x+2). Remaining denominator: (x-2). Vertical asymptote at x = 2 only. There is a hole at x = -2.
Q: Evaluate lim(x->0) 1/x^2.
A: From both sides, 1/x^2 goes to +infinity. So lim(x->0) 1/x^2 = +infinity.
Q: Evaluate lim(x->5-) 1/(x - 5) and lim(x->5+) 1/(x - 5).
A: As x approaches 5 from the left, (x - 5) is small and negative, so 1/(x - 5) goes to -infinity. From the right, (x - 5) is small and positive, so 1/(x - 5) goes to +infinity.
Q: Where does f(x) = x^3 sec(x) have a vertical asymptote on [0, pi]?
A: sec(x) = 1/cos(x). cos(x) = 0 at x = pi/2. Vertical asymptote at x = pi/2.
Q: Evaluate lim(x->(pi/2)-) x^3 sec(x) and lim(x->(pi/2)+) x^3 sec(x).
A: x^3 is positive near pi/2. As x->(pi/2)-, cos(x) is positive and approaching 0, so sec(x) goes to +infinity, giving limit = +infinity. As x->(pi/2)+, cos(x) is negative and approaching 0, so sec(x) goes to -infinity, giving limit = -infinity.
Limits are the foundation of everything that follows in Calculus I. The formal definition of the derivative is a limit of the difference quotient from Section 1.4. Continuity (coming next) is defined entirely in terms of limits. Vertical asymptotes will reappear when you sketch curves and analyse function behaviour in later chapters.
Limits, one-sided limit, two-sided limit, left-hand limit, right-hand limit, vertical asymptote, infinite limit, DNE, does not exist, rational function asymptotes, holes vs asymptotes, removable discontinuity, piecewise function limits, sign analysis, sec x, Calculus I, MTH 132, MSU, Michigan State