Difficulty: Foundational | Prerequisites: College algebra, basic function notation
Limits are the foundation of calculus. Every major concept in this course, from derivatives to integrals, is built on the idea of what a function approaches as the input gets close to some value. Continuity tells you whether a function behaves "nicely" at a point (no jumps, holes or blow-ups). You need limits and continuity before you can define the derivative or make sense of anything that follows. If you skipped straight to differentiation and feel lost, come back here.
A limit describes the value a function approaches as x gets close to a target, even if the function never reaches that value. Continuity means the function's limit and its actual value at a point agree, with no gaps or jumps. Asymptotes and holes describe what happens when continuity breaks down: the function either blows up (vertical asymptote), levels off (horizontal asymptote) or has a single missing point (hole).
Limit
The value that f(x) approaches as x approaches a particular number c. Written lim(x→c) f(x) = L. In simple terms, it is the answer to "where is this function headed?" even if it never arrives.
One-sided limit
The value f(x) approaches from only the left (x→c−) or only the right (x→c+). Think of it as checking the function's behaviour from one direction only. The two-sided limit exists only when both one-sided limits agree.
Limit at infinity
The value f(x) approaches as x grows without bound (x→∞) or shrinks without bound (x→−∞). In simple terms, this tells you the function's long-run behaviour, what the graph settles toward far to the right or far to the left.
Continuity
A function f is continuous at x = c when three conditions hold: f(c) is defined, lim(x→c) f(x) exists, and lim(x→c) f(x) = f(c). Think of it as: you can draw the graph through that point without lifting your pen.
Discontinuity
Any point where at least one of the three continuity conditions fails. The function jumps, has a hole, or blows up.
Horizontal asymptote
A horizontal line y = L that the graph approaches as x→∞ or x→−∞. Found by evaluating lim(x→±∞) f(x). In simple terms, it is the ceiling or floor the curve flattens toward.
Vertical asymptote
A vertical line x = a where the function grows without bound (the output shoots to ∞ or −∞). Found where the denominator equals zero and the factor does not cancel.
Hole (removable discontinuity)
A single missing point on the graph. It occurs when a factor cancels from both the numerator and denominator of a rational function. The limit exists at that x-value, but the function is either undefined there or defined as something different.
DNE (Does Not Exist)
Used when a limit fails to settle on a single finite value. Common reasons: the left-hand and right-hand limits disagree, or the function oscillates, or it diverges to ±∞.
Direct substitution: Plug the target value straight into f(x). If you get a real number, that is the limit. Works whenever the function is continuous at that point.
Example: lim(x→15) (−30) = −30. The function is a constant, so substitution gives −30 regardless of x.
Factor and cancel: When substitution gives 0/0, factor the numerator and denominator, cancel the common factor, then substitute again.
Example: lim(x→2) (x−2)/((x−2)(x+2)). Cancel (x−2) to get 1/(x+2), then substitute x = 2 to get 1/4.
Rationalise: When substitution gives 0/0 and a square root is involved, multiply top and bottom by the conjugate.
Example: lim(x→25) (x−25)/(√x − 5). Multiply by (√x + 5)/(√x + 5) to get √x + 5, then substitute to get 10.
Limits at infinity (rational functions): Compare the degree of the numerator (n) to the degree of the denominator (d).
n < d: the limit is 0.
n = d: the limit is the ratio of leading coefficients.
n > d: the limit is ±∞ (no horizontal asymptote).
Example: lim(x→∞) (3x² − 7x + 1)/(6x² + 8). Degrees equal, so the limit is 3/6 = 1/2.
Limits at −∞ for polynomials: The leading term dominates. If the leading term goes to +∞, the limit is +∞, and vice versa.
Example: lim(x→−∞) (−5x⁵ + 6x − 10). The leading term −5x⁵ with x→−∞ gives (−5)(−∞)⁵ = (−5)(−∞) = +∞.
One-sided limits: Approach from one direction only. Watch the sign of the denominator as x approaches the target from the specified side.
Example: lim(x→−4+) (x−3)/(x+4). As x→−4 from the right, the numerator → −7 (negative) and the denominator → 0+ (small positive), so the ratio → −∞.
To show f(x) is continuous at x = c, verify all three:
f(c) is defined.
lim(x→c) f(x) exists (left-hand limit = right-hand limit).
lim(x→c) f(x) = f(c).
If any condition fails, the function is not continuous at that point. State which condition breaks.
Example: f(x) = {2x−1 if x<3, −5 if x=3, x²−4 if x>3}. Check at x = 3:
f(3) = −5 (defined).
Left-hand limit: 2(3)−1 = 5. Right-hand limit: 3²−4 = 5. So the two-sided limit is 5.
But 5 ≠ −5, so condition 3 fails. Not continuous.
For a piecewise function with an unknown constant A, set the two pieces equal at the boundary to force continuity.
Example: f(x) = {2x²−3x+1 if x<−2, Ax−7 if x≥−2}. At x = −2: left piece = 2(4)+6+1 = 15. Set Ax−7 = 15 at x = −2: −2A−7 = 15, so A = −11.
Vertical asymptotes: Set the denominator equal to zero. If the factor does not cancel with the numerator, that x-value is a vertical asymptote.
Horizontal asymptotes: Evaluate the limit as x→±∞ using the degree rules above.
Holes: If a factor cancels from numerator and denominator, the function has a hole at that x-value. To find the y-coordinate of the hole, substitute into the simplified function.
Example: f(x) = (x−5)/(x²−4x−5). Factor: (x−5)/((x−5)(x+1)). The (x−5) cancels, leaving 1/(x+1). Hole at x = 5: y = 1/6. Vertical asymptote at x = −1. Horizontal asymptote: degree of numerator < degree of denominator, so y = 0.
\lim_{x \to c} f(x) = L \quad \text{(limit definition)}\text{Continuity checklist: } f(c) \text{ defined, } \lim_{x \to c} f(x) \text{ exists, } \lim_{x \to c} f(x) = f(c)\text{Horizontal asymptote rules for } \frac{a_n x^n + \cdots}{b_d x^d + \cdots}:n < d: HA is y = 0
n = d: HA is y = a_n / b_d
n > d: no HA
A = P e^{rt} \quad \text{(continuous compounding)}A = P\left(1 + \frac{r}{n}\right)^{nt} \quad \text{(compounding } n \text{ times per year)}The average rate of change formula (change in output divided by change in input) is the same idea as calculating speed from distance and time, or working out how fast petrol prices rose per year. Continuous compounding shows up in bank interest calculations and population growth models. Limits at infinity are how economists describe long-run equilibrium prices.
Students often think lim(x→c) f(x) must equal f(c). It does not have to. The limit describes where the function is headed, not where it is. They are equal only when the function is continuous at c.
Students confuse "the limit is infinity" with "the limit exists." When a function goes to ∞ or −∞, the limit does not exist as a finite number. Writing "= ∞" is shorthand for divergence, not an actual value.
When asked for horizontal asymptotes of y = x²/(x−1), students sometimes set the denominator to zero and call x = 1 the horizontal asymptote. That gives the vertical asymptote. Horizontal asymptotes come from limits at infinity.
A common error is to cancel (x−2) from a fraction like (x−2)/((x−2)(x+2)) without noting that x = 2 is excluded from the domain. The cancellation simplifies the expression, but x = 2 is a hole, not a regular point.
⚠️ Expect at least one piecewise continuity problem. Know the three-condition checklist cold.
⚠️ "Find the value of A that makes f(x) continuous" is a recurring question. Set the two pieces equal at the boundary.
⚠️ Asymptote and hole problems require you to show limit work. The answer key warns: "Failure to use limits will result in a zero."
⚠️ The limit of a constant function is just that constant, regardless of where x is headed. Do not multiply the constant by the target value.
True or false: If lim(x→c) f(x) = 5 and f(c) = 3, the function is continuous at c. (False, condition 3 fails.)
Fill in the blank: When the degree of the numerator equals the degree of the denominator, the horizontal asymptote is y = ______. (The ratio of leading coefficients.)
True or false: lim(x→15) (−30) = −450. (False, it is −30. A constant function's limit is the constant itself.)
Fill in the blank: A hole in a rational function occurs when a factor ______ from both numerator and denominator. (Cancels.)
True or false: If the left-hand limit is 3 and the right-hand limit is −3, the two-sided limit is 0. (False, the two-sided limit does not exist.)
Q: Evaluate lim(x→7) (2x)/(x+1).
A: Direct substitution. 2(7)/(7+1) = 14/8 = 7/4.
Q: Evaluate lim(x→25) (x−25)/(√x − 5).
A: Substitution gives 0/0. Multiply top and bottom by (√x + 5). The numerator becomes (x−25) and the denominator becomes (√x − 5)(√x + 5) = x − 25. Cancel (x−25) to get √x + 5. Substitute x = 25: √25 + 5 = 10.
Q: Evaluate lim(x→∞) (3x² − 7x + 1)/(6x² + 8).
A: Degrees are equal (both 2). Limit = 3/6 = 1/2.
Q: Evaluate lim(x→−∞) (−5x⁵ + 6x − 10).
A: Leading term dominates. −5(−∞)⁵ = −5(−∞) = +∞. The limit is ∞.
Q: Evaluate lim(x→−4+) (x−3)/(x+4).
A: As x→−4 from the right, numerator → −7 and denominator → 0+. A negative divided by a tiny positive is −∞.
Q: Find the value of A that makes f(x) = {2x²−3x+1 for x<−2, Ax−7 for x≥−2} continuous on all reals.
A: At x = −2: left piece = 2(4)+6+1 = 15. Right piece at x = −2: −2A−7. Set equal: −2A−7 = 15, so −2A = 22, A = −11.
Q: For f(x) = (x−5)/(x²−4x−5), find the horizontal asymptote, vertical asymptote, and any holes.
A: Factor denominator: (x−5)(x+1). The (x−5) cancels, leaving 1/(x+1). Hole at x = 5 with y = 1/6 (substitute into simplified function). Vertical asymptote at x = −1. Horizontal asymptote: original function has degree 1 on top, degree 2 on bottom, so y = 0.
Q: How long for $3,000 to become $30,000 at 3.7% compounded continuously?
A: Use A = Pe^(rt). 30000 = 3000e^(0.037t). Divide: 10 = e^(0.037t). Take ln: ln 10 = 0.037t. t = ln(10)/0.037 ≈ 62 years.
Q: Find the balance when $800 is invested at 4.6% compounded monthly for 10 years.
A: A = 800(1 + 0.046/12)^(12×10) = 800(1.003833...)^120 ≈ $1,266.15.
Limits are the gateway to the derivative: the definition of the derivative is itself a limit (lim(h→0) [f(x+h)−f(x)]/h). You will use continuity to decide where differentiation rules apply. Horizontal asymptotes reappear in curve sketching when you analyse long-run behaviour. The definite integral is also defined as a limit of Riemann sums, so limits thread through the entire course.
limit, one-sided limit, left-hand limit, right-hand limit, limit at infinity, continuity, discontinuity, removable discontinuity, hole, vertical asymptote, horizontal asymptote, piecewise function, indeterminate form, 0/0, direct substitution, factor and cancel, rationalise, conjugate, continuous compounding, compound interest, DNE, does not exist, MAC 2233, calculus for business, FIU