Difficulty: Foundational | Prerequisites: Algebra and basic function notation
Limits, continuity, and rates of change are the gateway topics of calculus. Everything that follows (derivatives, integrals, optimisation) rests on the idea that you can describe what a function does as its input creeps toward a particular value, or as it grows without bound. If these concepts are shaky, the rest of the course will feel like guesswork. You should already be comfortable with factoring polynomials, simplifying rational expressions, and reading piecewise function notation.
A limit tells you the value a function approaches (not necessarily reaches) as x nears a target. A function is continuous at a point when the limit and the function value agree. Horizontal asymptotes describe a function's long-run behaviour. The average rate of change between two points is just slope: rise over run.
Limit
The value that f(x) approaches as x gets arbitrarily close to a particular number. The function does not need to equal this value at the point itself.
Think of it as: where the function is headed, not necessarily where it lands.
Continuity
A function f(x) is continuous at x = a when three conditions hold: f(a) exists, the limit of f(x) as x approaches a exists, and the limit equals f(a). No gaps, no jumps, no holes.
In simple terms, you can draw the graph through that point without lifting your pen.
Piecewise function
A function defined by different rules on different intervals of its domain. Each piece has its own formula, and you pick which one to use based on the value of x.
Think of it as: a function with multiple personalities, one per interval.
Horizontal asymptote
A horizontal line y = L that the graph of f(x) approaches as x goes to positive or negative infinity. It describes the function's long-run behaviour.
In simple terms, it is the value the outputs settle toward when x gets very large (or very negative).
Average rate of change
The ratio of the change in f(x) to the change in x over an interval [a, b]. Geometrically, it is the slope of the secant line connecting (a, f(a)) and (b, f(b)).
Think of it as: the slope between two points on the curve.
When direct substitution gives 0/0 (an indeterminate form), factor the numerator and denominator, cancel the common factor, then substitute again.
Example: lim(x→5) of (x² − 25)/(x − 5). Factor the numerator as (x − 5)(x + 5), cancel (x − 5), leaving x + 5. Substitute x = 5 to get 10.
The cancelled factor tells you where the hole in the graph is, not the limit itself. The limit still exists even though the function is undefined at that point.
Compare the degree of the numerator (n) and the degree of the denominator (m).
If n < m, the horizontal asymptote is y = 0.
If n = m, the horizontal asymptote is y = (leading coefficient of numerator) / (leading coefficient of denominator).
If n > m, there is no horizontal asymptote.
Example: f(x) = (7x³ + 5x² − 1)/(3x³ + 5). Both numerator and denominator have degree 3, so the asymptote is y = 7/3.
For a piecewise function to be continuous at the boundary point x = c, the left-hand piece and the right-hand piece must give the same output at x = c.
Set the two expressions equal at the boundary and solve for the unknown constant.
Example: f(x) = {3x + 1 if x < 2, Ax² + 7 if x ≥ 2}. At x = 2: left piece gives 3(2) + 1 = 7. Right piece gives A(4) + 7. Set 4A + 7 = 7, so A = 0.
The average rate of change of f(x) from x = a to x = b is [f(b) − f(a)] / (b − a).
This is the slope of the secant line through the two points.
Example: gas cost was $2.80 in 2007 and $3.50 in 2014. Average rate of change = (3.50 − 2.80)/(2014 − 2007) = 0.70/7 = $0.10 per year.
Limit by factoring (difference of squares)
lim(x→a) (x² − a²)/(x − a) = lim(x→a) (x + a) = 2a
Horizontal asymptote rule
For f(x) = (aₙxⁿ + ...)/(bₘxᵐ + ...): if n = m, asymptote is y = aₙ/bₘ; if n < m, asymptote is y = 0; if n > m, no horizontal asymptote.
Average rate of change
[f(b) − f(a)] / (b − a)
Piecewise continuity condition
At a boundary x = c: set the left-hand expression equal to the right-hand expression and solve for the unknown.
Students often think that if direct substitution gives 0/0, the limit does not exist. It does not mean that. 0/0 is indeterminate, which means you need to do more work (factor, simplify, rationalise) before you can evaluate the limit.
Students sometimes confuse the value of a function at a point with the limit at that point. A function can be undefined at x = 5 and still have a perfectly well-defined limit there.
When finding horizontal asymptotes, students frequently compare individual terms instead of leading terms. Only the highest-degree terms in the numerator and denominator matter as x grows large.
With piecewise continuity, students occasionally check only that both pieces are defined at the boundary rather than checking that they produce the same output there. Being defined at the boundary is necessary but not sufficient.
⚠️ Limit evaluation by factoring is one of the most commonly tested skills on the final. Know your difference of squares and common factor patterns cold.
⚠️ Horizontal asymptote questions almost always involve comparing the degrees of numerator and denominator. Memorise the three cases.
⚠️ Piecewise continuity problems appear regularly. The method is always the same: plug the boundary value into both pieces, set them equal, solve.
⚠️ Average rate of change questions can be disguised as word problems (cost over time, distance over time). Recognise that you are being asked for a slope.
Limits underpin how engineers and economists model systems at extreme values or near breaking points. Average rate of change is how businesses compute growth rates, inflation, or cost trends from one period to another.
True or False: If f(3) is undefined, then lim(x→3) f(x) does not exist.
Fill in the blank: For f(x) = (5x⁴ + 2)/(5x⁴ − 1), the horizontal asymptote is y = ___.
True or False: A piecewise function is continuous at a boundary if both pieces are defined there.
Fill in the blank: The average rate of change formula is ___.
Q: Evaluate lim(x→5) (x² − 25)/(x − 5).
A: Factor the numerator as (x − 5)(x + 5), cancel (x − 5), substitute x = 5 into (x + 5) to get 10.
Q: What is the horizontal asymptote of f(x) = (7x³ + 5x² − 1)/(3x³ + 5)?
A: Both numerator and denominator have degree 3. The asymptote is y = 7/3.
Q: Find A so that f(x) = {3x + 1 if x < 2, Ax² + 7 if x ≥ 2} is continuous everywhere.
A: At x = 2, the left piece gives 7 and the right piece gives 4A + 7. Setting 4A + 7 = 7 gives A = 0.
Q: Gas cost $2.80 in 2007 and $3.50 in 2014. What is the average rate of change per year?
A: (3.50 − 2.80)/(2014 − 2007) = 0.70/7 = $0.10 per year.
Q: What are the three cases for horizontal asymptotes of a rational function?
A: If the numerator's degree is less than the denominator's, y = 0. If they are equal, y = ratio of leading coefficients. If the numerator's degree is greater, there is no horizontal asymptote.
Limits are the foundation of the derivative, which is itself defined as a limit of the difference quotient. The average rate of change you compute here becomes the instantaneous rate of change (the derivative) when you shrink the interval to zero. Horizontal asymptotes reappear in optimisation and curve sketching, where you need to know what happens to a function at the extremes.
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