Difficulty: Introductory | Prerequisites: Basic algebra (factoring, simplifying rational expressions)
Tags: limits, continuity, polynomial functions, limits at infinity, rational functions, direct substitution, indeterminate forms, calculus for business, FIU
Limits are the gateway concept in calculus. Everything that follows (derivatives, integrals, optimisation) rests on understanding what a function does as its input approaches a particular value. This activity covers the mechanics: evaluating limits by direct substitution, handling indeterminate forms through factoring, and determining what polynomials and rational functions do as x heads towards positive or negative infinity. If you can do these reliably, the transition to derivatives will feel far less steep.
A limit describes the value a function approaches as x gets close to some number (or heads to infinity). For most polynomials, you can just plug in the number. When plugging in gives 0/0, factor and cancel first. For limits at infinity, only the highest-degree term matters.
Limit
The value that f(x) approaches as x gets arbitrarily close to a given number. Written as lim (x→a) f(x). In simple terms: what the y-value is heading towards as x sneaks up on some target.
Direct substitution
Evaluating a limit by plugging the target x-value straight into the function. Think of it as: the simplest case, where nothing breaks when you just substitute.
Indeterminate form (0/0)
The result you get when direct substitution produces 0 divided by 0. This does not mean the limit is zero or undefined; it means more work is needed (usually factoring). In simple terms: 0/0 is a signal to simplify, not an answer.
Continuity
A function is continuous at a point if the limit at that point equals the function's value there. Polynomials are continuous everywhere. Think of it as: you can draw the graph through that point without lifting your pen.
Polynomial function
A function built from terms of the form ax^n, where n is a non-negative integer. Examples: 3x^2 + 2x + 1, x^7 - 8x^3 + 2. In simple terms: the bread and butter of algebra, no fractions or roots of x involved.
Rational function
A function written as one polynomial divided by another. Example: (4x^2 - 1) / (5x^3 + 8). Think of it as: a fraction where both the top and bottom are polynomials.
Leading term
The term in a polynomial with the highest power of x. This term dominates the function's behaviour as x approaches infinity. In simple terms: when x is enormous, only the biggest power matters.
Limit at infinity
The value (or behaviour) of a function as x grows without bound in the positive or negative direction. Think of it as: what happens to y when x goes off the edge of your graph paper.
When a function is continuous at x = a, the limit equals f(a). Polynomials are continuous everywhere, so for any polynomial you can substitute directly.
Example: lim (x→3) (x^2 - x + 3) = 3^2 - 3 + 3 = 9
Example: lim (x→7) 20 = 20. A constant function always returns the same value, so its limit is just that constant, regardless of where x is heading.
When direct substitution gives a valid number (not 0/0), you are done. When it produces something like (4 - 16) / (2 - 4), that is just arithmetic, and you carry it through.
Example: lim (x→4) (x - 16) / (sqrt(x) - 4) = (4 - 16) / (sqrt(4) - 4) = (-12) / (2 - 4) = (-12) / (-2) = 6
If substituting gives 0/0, factor the numerator and denominator, cancel the common factor, then substitute again.
Example: lim (x→1) (x^2 - 1) / (x - 1)
Substituting x = 1 gives (1 - 1) / (1 - 1) = 0/0. This is indeterminate.
Factor: x^2 - 1 = (x + 1)(x - 1)
Cancel: (x + 1)(x - 1) / (x - 1) = (x + 1), valid for x ≠ 1
Now substitute: lim (x→1) (x + 1) = 2
The cancellation is legal because limits care about what happens near x = 1, not at x = 1 itself.
For any polynomial, only the leading term (highest power of x) determines what happens as x→∞ or x→−∞. The other terms become negligible.
Example: lim (x→−∞) (4x^7 - 8x^3 + 2) = lim (x→−∞) 4x^7 = −∞
The leading term is 4x^7. An odd power of a negative number is negative, so 4 times a hugely negative number is −∞.
Example: lim (x→∞) (5x^2 - 2x^4 - 10x + 1) = lim (x→∞) (-2x^4) = −∞
The leading term is -2x^4. An even power is always positive, so -2 times a huge positive number is −∞.
Example: lim (x→−∞) ((3x^3 - 1)(x^5 - 3))
Multiply the leading terms: 3x^3 times x^5 = 3x^8.
lim (x→−∞) 3x^8 = +∞. An even power of any large number is positive.
Compare the degree of the numerator (top) with the degree of the denominator (bottom).
Degree of numerator < degree of denominator: the limit is 0.
Example: lim (x→−∞) (4x^2 - 1) / (5x^3 + 8) = 0. The bottom grows faster, so the fraction shrinks to zero.
Degree of numerator = degree of denominator: the limit is the ratio of the leading coefficients.
Example: if you had (6x^3 + 1) / (2x^3 - 5), the limit would be 6/2 = 3.
Degree of numerator > degree of denominator: the limit is ±∞. The sign depends on the leading coefficients and the direction of x.
Example: lim (x→−∞) (4x^3 - 1) / (5x^2 + 8). The top has degree 3, the bottom degree 2. Simplify to lim (x→−∞) (4x/5) = −∞.
\lim_{x \to a} f(x) = f(a) \quad \text{(direct substitution, when } f \text{ is continuous at } a\text{)}\lim_{x \to a} \frac{f(x)}{g(x)} = \frac{0}{0} \implies \text{factor, cancel, then re-substitute}\lim_{x \to \pm\infty} (a_n x^n + a_{n-1} x^{n-1} + \cdots) = \lim_{x \to \pm\infty} a_n x^n\lim_{x \to \pm\infty} \frac{a_m x^m + \cdots}{b_n x^n + \cdots} = \begin{cases} 0 & m < n \\ \frac{a_m}{b_n} & m = n \\ \pm\infty & m > n \end{cases}Difference of squares (common factoring pattern for 0/0):
x^2 - 1 = (x+1)(x-1)Limits at infinity are how economists and analysts describe long-run behaviour of cost, revenue, and profit functions. If your average cost function is a rational expression, its limit as quantity grows tells you the cost floor your business approaches at scale. Limits at finite points underpin marginal analysis: the derivative (coming next in the course) is defined as a limit, so every marginal cost or marginal revenue calculation you will do later is a limit in disguise.
"0/0 means the limit is 0 (or undefined)." It does not. 0/0 is an indeterminate form, which means the limit could be any finite number, or could be infinite. You need to do more algebra to find out.
"A limit at infinity must be a number." No. Polynomials blow up to ±∞, so the limit is not a finite value. Saying the limit "equals" ∞ is shorthand for "grows without bound."
"You can ignore the sign of the leading coefficient." The sign matters. lim (x→∞) (-2x^4) is −∞, not +∞, because of the negative coefficient.
"For rational functions at infinity, just look at the biggest term overall." You must compare the degree of the numerator with the degree of the denominator separately. The ratio of those leading terms is what determines the answer.
⚠️ The 0/0 indeterminate form is a near-certain exam topic. Expect at least one problem where direct substitution fails and factoring is required.
⚠️ Limits at infinity for rational functions: know the three cases (degree top < bottom, equal, top > bottom) cold. These appear as quick multiple-choice questions.
⚠️ Sign errors on limits at infinity are the most common point-loser. Track the sign of the leading coefficient and whether the power is odd or even separately.
⚠️ The constant-function limit (lim of a constant = that constant) looks trivial but gets tested to see if students overthink it.
True or false: lim (x→5) 12 = 5. Answer: False. The limit of a constant is the constant itself, so the answer is 12.
Fill in the blank: When direct substitution gives 0/0, the next step is to ______ the expression. Answer: Factor (and cancel the common factor).
True or false: lim (x→∞) (3x^2 + 100x) = lim (x→∞) 100x because 100 > 3. Answer: False. The highest-degree term dominates, so the limit equals lim (x→∞) 3x^2.
Fill in the blank: For a rational function where the numerator's degree is less than the denominator's degree, the limit as x→±∞ is ______. Answer: 0.
True or false: lim (x→1) (x^2 - 1)/(x - 1) is undefined because the function is undefined at x = 1. Answer: False. The limit is 2. A limit describes what the function approaches, not its value at that point.
Q: Evaluate lim (x→2) (x^2 + 3x - 1).
A: Substitute directly: 2^2 + 3(2) - 1 = 4 + 6 - 1 = 9.
Q: Evaluate lim (x→3) (x^2 - 9) / (x - 3).
A: Direct substitution gives 0/0. Factor the numerator: x^2 - 9 = (x + 3)(x - 3). Cancel (x - 3). The limit becomes lim (x→3) (x + 3) = 6.
Q: Evaluate lim (x→∞) (7x^5 - 3x^2 + 1).
A: The leading term is 7x^5. As x→∞, 7x^5→+∞. The limit is +∞.
Q: Evaluate lim (x→−∞) (6x^3 + 2) / (3x^3 - x).
A: Numerator and denominator have the same degree (3). The limit is the ratio of leading coefficients: 6/3 = 2.
Q: Evaluate lim (x→−∞) (x^2 + 1) / (x^5 - 4).
A: Degree of numerator (2) < degree of denominator (5). The limit is 0.
This material feeds directly into derivatives. The definition of a derivative is a limit of the form lim (h→0) (f(x+h) - f(x)) / h, which is exactly the 0/0 indeterminate form you practised here. If limits feel mechanical now, that is the point: they should be automatic by the time you meet derivatives.
Limits at infinity connect to asymptotic behaviour and graphing. When you sketch rational functions later, the horizontal asymptote is the limit as x→±∞, and you will use the degree-comparison rules from this activity.
Continuity ties into the Intermediate Value Theorem, which the course may use to prove that certain business equations (break-even, equilibrium) have solutions.
limits, limit evaluation, direct substitution, continuity, polynomial limits, rational function limits, indeterminate form, 0/0, factoring limits, leading term, degree comparison, horizontal asymptote, limits at infinity, end behaviour, calculus for business, FIU calculus, limit rules, limit of a constant, difference of squares factoring