Difficulty: Introductory to Intermediate | Prerequisites: College algebra, familiarity with function notation.
This material sits at the front of MAC 2233 and threads through the entire course. Limits are the foundation that derivatives and integrals are built on. Continuity tells you whether a function behaves predictably at a point, and asymptotes describe what happens at the extremes. If you skipped or struggled with limits early in the semester, nearly every later topic will feel harder than it needs to. Average rate of change is the discrete cousin of the derivative, and recognising the connection between the two is a common exam question.
TL;DR: Limits let you find a function's value (or behaviour) at points where plugging in directly fails. Continuity means no breaks, jumps, or holes. Asymptotes are lines the graph approaches but never crosses. Average rate of change is slope between two points, and it bridges algebra into calculus.
Limit
The value a function f(x) approaches as x gets arbitrarily close to a particular number c. Written lim (x→c) f(x). Think of it as: "What does the output get close to as the input sneaks up on c?"
One-sided limit
The value f(x) approaches from only the left (x→c⁻) or only the right (x→c⁺). In simple terms, you are checking the function's behaviour from one direction only.
Continuity at a point
A function f is continuous at x = c when three conditions hold: f(c) exists, lim (x→c) f(x) exists, and f(c) equals that limit. Think of it as: you can draw through that point without lifting your pen.
Removable discontinuity (hole)
A point where the limit exists but the function is either undefined or disagrees with the limit. The graph has a literal hole at that coordinate. In simple terms, there is a "missing dot" you could fill in to make the function continuous.
Vertical asymptote
A vertical line x = a where the function grows without bound (toward +∞ or −∞) as x approaches a. The denominator equals zero at x = a and the factor does not cancel. Think of it as the x-value the function can never touch, where it shoots off to infinity.
Horizontal asymptote
A horizontal line y = L that the graph approaches as x→∞ or x→−∞. It describes the function's long-run behaviour. In simple terms, this is the value the output settles toward when x gets very large (positive or negative).
Average rate of change
The slope of the secant line between two points on a function: [f(b) − f(a)] / (b − a). Think of it as the "overall speed" between two moments, ignoring what happened in between.
Difference of squares
The algebraic identity a² − b² = (a + b)(a − b). This factoring pattern appears constantly in limit problems where direct substitution gives 0/0.
Formula: (f(b) − f(a)) / (b − a).
This is the slope of the straight line (secant line) connecting two points on a curve.
Example: gasoline cost rises from $2.84 in 2007 to $3.44 in 2014. Average rate of change = (3.44 − 2.84) / (2014 − 2007) = 0.60 / 7 ≈ $0.09 per year.
Units matter. The result carries the units of the output divided by the units of the input (here, dollars per year).
When direct substitution gives 0/0, factor the numerator and denominator to cancel the common factor.
Example: lim (x→5) of (x² − 25)/(x − 5). Factor the numerator as (x + 5)(x − 5). Cancel (x − 5). You are left with lim (x→5) of (x + 5) = 10.
The cancelled factor tells you there is a hole at x = 5, not a vertical asymptote.
For a piecewise function to be continuous at the boundary x = c, the left piece and the right piece must agree at x = c.
Method: evaluate the left piece at x = c, evaluate the right piece at x = c, set them equal, and solve for the unknown constant.
Example: f(x) = { 3x + 1 if x < 2; Ax + 7 if x ≥ 2 }. Left piece at x = 2 gives 3(2) + 1 = 7. Right piece at x = 2 gives 2A + 7. Set 7 = 2A + 7, so A = 0.
Compare the degree of the numerator (n) with the degree of the denominator (d).
n < d: horizontal asymptote at y = 0.
n = d: horizontal asymptote at y = (leading coefficient of numerator) / (leading coefficient of denominator).
n > d: no horizontal asymptote.
Example: y = (7x³ − 5x² + 3x) / (x³ + 5). Degrees are equal (both 3), so y = 7/1 = 7.
Set the denominator equal to zero. Each solution is either a vertical asymptote or a hole.
If the factor cancels with one in the numerator, it is a hole. If it does not cancel, it is a vertical asymptote.
To find the y-coordinate of a hole, cancel the common factor and substitute the x-value into the simplified function.
Example: f(x) = (x² − 4x − 5) / (x − 5). Factor numerator: (x − 5)(x + 1). The (x − 5) cancels, so x = 5 is a hole (not an asymptote). Simplified function is x + 1, so the hole is at (5, 6). Since all factors cancel, there is no vertical asymptote. The degree of numerator (2) exceeds the degree of denominator (1), so there is no horizontal asymptote either.
\text{Average rate of change} = \frac{f(b) - f(a)}{b - a}\lim_{x \to c} f(x) = L\text{Horizontal asymptote (equal degrees): } y = \frac{a_n}{b_n}a^2 - b^2 = (a+b)(a-b)Average rate of change is the basis of every "how fast is it growing" question in business: revenue growth per quarter, cost increases per year, population change per decade. Asymptotes show up whenever a model has a ceiling (market saturation) or a floor (minimum cost). Continuity matters in engineering and economics any time a sudden jump in a function would break the model or the system.
Students often think a limit must equal the function's value at that point. It does not. The limit describes what the function approaches, which can differ from f(c) or exist even when f(c) is undefined.
Students confuse holes with vertical asymptotes. If the zero in the denominator cancels with a factor in the numerator, it is a hole, not an asymptote.
When degrees are equal, students sometimes compare all coefficients rather than just the leading ones. Only the leading coefficients determine the horizontal asymptote.
Students sometimes believe a function cannot cross its horizontal asymptote. It can. The asymptote describes end behaviour (as x→±∞), not a barrier.
⚠️ Factoring to resolve 0/0 indeterminate forms is tested in almost every exam. Know your factoring patterns cold, especially difference of squares.
⚠️ Piecewise continuity problems ("find the value of A") appear regularly. The method is always the same: set the two pieces equal at the boundary.
⚠️ Horizontal asymptote questions require you to compare degrees quickly. Memorise the three cases.
⚠️ Hole vs. vertical asymptote is a classic exam distinction. If you can cancel the factor, it is a hole. If you cannot, it is a vertical asymptote.
True or False: If f(3) = 7, then lim (x→3) f(x) must also equal 7.
False. The function value and the limit can differ (or the limit might not exist).
Fill in the blank: When the degree of the numerator is less than the degree of the denominator, the horizontal asymptote is y = ___.
y = 0.
True or False: A factor that cancels in both numerator and denominator produces a vertical asymptote.
False. It produces a hole (removable discontinuity).
Fill in the blank: The average rate of change formula is ___ / ___.
[f(b) − f(a)] / [b − a].
Q: Evaluate lim (x→5) of (x² − 25) / (x − 5).
A: Factor the numerator as (x+5)(x−5). Cancel (x−5). The limit equals 5 + 5 = 10.
Q: The price of a commodity was $2.84 in 2007 and $3.44 in 2014. What is the average rate of change per year?
A: (3.44 − 2.84) / (2014 − 2007) = 0.60 / 7 ≈ $0.09 per year.
Q: For y = (7x³ − 5x² + 3x) / (x³ + 5), what is the horizontal asymptote?
A: Degrees are equal (both 3). Horizontal asymptote is y = 7/1 = 7.
Q: Given f(x) = { 3x + 1 for x < 2; Ax + 7 for x ≥ 2 }, find A so that f is continuous everywhere.
A: At x = 2, the left piece gives 3(2) + 1 = 7. The right piece gives 2A + 7. Set 7 = 2A + 7, so A = 0.
Q: For f(x) = (x² − 4x − 5) / (x − 5), identify any holes and vertical asymptotes.
A: Factor numerator: (x − 5)(x + 1). The (x − 5) cancels, so x = 5 is a hole at (5, 6). No vertical asymptote remains.
The limit is the conceptual engine behind derivatives (the instantaneous rate of change is a limit of the average rate of change as the interval shrinks to zero) and definite integrals (Riemann sums). Continuity is a prerequisite for the Intermediate Value Theorem and for most applied optimisation problems. Asymptotic behaviour feeds directly into curve-sketching questions.
limit, continuity, discontinuity, removable discontinuity, hole in graph, vertical asymptote, horizontal asymptote, average rate of change, secant line, indeterminate form, 0/0, factoring limits, difference of squares, piecewise function, degree comparison, end behaviour, MAC 2233, calculus for business, FIU