Difficulty: Foundational | Prerequisites: Algebra, basic trigonometry
Limits describe what value a function approaches as the input gets close to some number, even if the function never reaches that value. Continuity means the function has no breaks, jumps, or holes at a given point. These two ideas are the foundation of calculus: you cannot define a derivative without them, and nearly every exam will test whether you can evaluate limits and classify discontinuities quickly.
Limit
The value that f(x) approaches as x gets closer and closer to a particular number. You write it as lim (x→a) f(x) = L. Think of it as: "Where is the function heading?" even if it never arrives.
One-sided limit
The value f(x) approaches from only one direction. A left-hand limit (x→a⁻) comes from below; a right-hand limit (x→a⁺) comes from above. The two-sided limit exists only when both one-sided limits agree.
Continuity
A function is continuous at x = a when three things hold: f(a) is defined, the limit as x→a exists, and the limit equals f(a). In simple terms, you can draw through the point without lifting your pen.
Removable discontinuity
A hole in the graph. The limit exists at that point, but the function value either does not exist or does not match the limit. You can "fix" it by redefining f(a) to equal the limit.
Jump discontinuity
The left-hand limit and the right-hand limit both exist but are not equal. The graph has a visible break, like a step.
Vertical asymptote (VA)
A vertical line x = c where the function grows without bound (toward +∞ or −∞). This happens when the denominator of a rational function equals zero but the numerator does not (after cancelling common factors).
Horizontal asymptote (HA)
The value y = L that f(x) approaches as x→∞ or x→−∞. It describes the long-run behaviour of the function. For a rational function, compare the degrees of numerator and denominator to find it.
The first thing to try. Plug the target value straight into the function.
If you get a real number, that is the limit. Done.
Example: lim (x→2) (x² − 1)/(x³ + 2) = (4 − 1)/(8 + 2) = 3/10.
The limit of a constant is the constant itself, regardless of what x does.
Example: lim (x→e) 7 = 7.
If direct substitution gives 0/0, factor the numerator and denominator, cancel the common factor, then substitute again.
Example: lim (x→−1) (x + 1)/(x² − 5x − 6). Factor the denominator as (x − 6)(x + 1). Cancel (x + 1) to get 1/(x − 6). Substitute x = −1: result is 1/(−7) = −1/7.
Example: lim (x→1) (x − 1)/(x³ − x² + x − 1). Factor the denominator as (x − 1)(x² + 1). Cancel (x − 1) to get 1/(x² + 1). Substitute x = 1: result is 1/2.
lim (x→0) sin(kx)/x = k. This comes from the foundational result lim (x→0) sin(x)/x = 1.
Example: lim (x→0) sin(6x)/x = 6.
lim (x→0) tan(x) = 0 (just plug in: tan(0) = 0).
f(a) is defined (the point exists on the graph).
lim (x→a) f(x) exists (the left-hand and right-hand limits agree).
lim (x→a) f(x) = f(a) (the limit equals the function value).
If any condition fails, the function is discontinuous at that point.
Set the two pieces equal at the boundary. For example, given f(x) = 3x³ for x ≤ 1 and f(x) = ax + 5 for x > 1, set 3(1)³ = a(1) + 5, which gives 3 = a + 5, so a = −2.
Removable (hole): The limit exists, but f(a) is either undefined or does not equal the limit. You can fix it by redefining f(a). Example from the review: the graph at x = 4 has lim (x→4) f(x) = 0, but f(4) = DNE. Assigning f(4) = 0 removes the discontinuity.
Jump: The left-hand and right-hand limits both exist but differ. The graph has a step. Example from the review: the graph at x = 2 has a jump discontinuity.
Infinite (vertical asymptote): The function blows up to ±∞. The limit does not exist in the finite sense.
When given a graph, check each point systematically:
At x = −4: lim (x→−4⁺) f(x) = −2, f(−4) = 2 (filled dot). The limit from the left does not match the function value, so not continuous.
At x = 0: The graph passes through without a break, so continuous.
At x = 4: The limit is 0 but f(4) is undefined (open circle with no filled dot). Not continuous, but removable.
A vertical asymptote occurs at x = c when the denominator is zero and the numerator is not (after all common factors are cancelled).
To prove it, show that the one-sided limits go to ±∞.
Worked example: f(x) = (x² + 7x + 12)/(x² + 5x + 6). Factor: (x + 4)(x + 3) / (x + 3)(x + 2). Cancel (x + 3). The simplified function is (x + 4)/(x + 2). The denominator is zero at x = −2 (and the numerator is not), so there is a vertical asymptote at x = −2. Note that x = −3 is a removable discontinuity (a hole), not a vertical asymptote.
To confirm: lim (x→−2⁺) f(x) = +∞ and lim (x→−2⁻) f(x) = −∞.
Compare the degree of the numerator (n) with the degree of the denominator (d).
If n < d: HA at y = 0. The denominator grows faster and pulls the fraction toward zero.
If n = d: HA at y = (leading coefficient of numerator)/(leading coefficient of denominator).
If n > d: No horizontal asymptote (the function grows without bound). A useful mnemonic from the review: "BOBO BATNA EATS DC" (Bigger On Bottom: zero; Bottom And Top the same: ratio of leading coefficients; Bigger On Top: no HA).
Worked example: f(x) = (2x + 5)/(3x² + x + 2). The numerator has degree 1, the denominator has degree 2. Since 1 < 2, the horizontal asymptote is y = 0. Confirm: lim (x→∞) f(x) = 0 and lim (x→−∞) f(x) = 0.
Students often assume that if the denominator is zero, the function has a vertical asymptote. This is not always the case. If the numerator is also zero at the same point (a common factor), you may have a removable discontinuity (hole) instead. Always factor and cancel first.
Students confuse "the limit does not exist" with "the function is not defined." These are different. A limit can exist even when f(a) is undefined (that is exactly what a removable discontinuity is). Conversely, f(a) can be defined while the limit fails to exist (as with a jump).
When finding horizontal asymptotes, students sometimes divide every term by x instead of by the highest power of x in the denominator. Stick to the degree-comparison shortcut: it is faster and less error-prone.
Students sometimes think a function cannot cross its horizontal asymptote. It can. The HA describes end behaviour (what happens as x→±∞), not behaviour in the middle of the graph.
⚠️ Expect at least one limit that requires factoring and cancelling before substitution. If you get 0/0, that is your cue to factor.
⚠️ The special limit lim (x→0) sin(kx)/x = k appears regularly. Memorise the pattern.
⚠️ Graph-reading questions will ask you to classify discontinuities (removable, jump, infinite) and to state what value would make the function continuous. Know all three conditions for continuity.
⚠️ For asymptotes, you will likely need to show the limit-based proof, not just state the answer. Write out lim (x→c⁺) and lim (x→c⁻) for vertical asymptotes, and lim (x→±∞) for horizontal asymptotes.
⚠️ Average rate of change = (f(b) − f(a))/(b − a). This is the slope of the secant line. Example from the review: f(x) = −x³ − 3 on [0, 6] gives (−219 − (−3))/(6 − 0) = −216/6 = −36.
True or false: If f(3) = 5, then lim (x→3) f(x) = 5. (False. The function value and the limit can differ.)
Fill in the blank: lim (x→0) sin(4x)/x = ___. (4)
True or false: A vertical asymptote at x = 2 means f(2) is defined. (False. The function is undefined at a vertical asymptote.)
Fill in the blank: A discontinuity where the limit exists but does not equal f(a) is called a ___ discontinuity. (removable)
True or false: If the degree of the numerator is less than the degree of the denominator, the horizontal asymptote is y = 0. (True)
Q: Evaluate lim (x→−1) (x + 1)/(x² − 5x − 6).
A: Factor the denominator as (x − 6)(x + 1). Cancel (x + 1). Substitute x = −1 into 1/(x − 6) to get 1/(−7) = −1/7.
Q: Given f(x) = { 3x³ if x ≤ 1, ax + 5 if x > 1 }, find the value of a that makes f continuous.
A: Set the left piece equal to the right piece at x = 1. 3(1)³ = a(1) + 5 gives 3 = a + 5, so a = −2.
Q: Find the vertical asymptote(s) of f(x) = (x² + 7x + 12)/(x² + 5x + 6). Prove with limits.
A: Factor: (x + 4)(x + 3)/((x + 3)(x + 2)). Cancel (x + 3). The simplified form is (x + 4)/(x + 2). The denominator is zero at x = −2, so VA at x = −2. Proof: lim (x→−2⁺) f(x) = +∞ and lim (x→−2⁻) f(x) = −∞.
Q: Find the horizontal asymptote of f(x) = (2x + 5)/(3x² + x + 2). Prove with limits.
A: Degree of numerator (1) < degree of denominator (2), so HA at y = 0. Proof: lim (x→∞) f(x) = 0 and lim (x→−∞) f(x) = 0.
Q: Calculate the average rate of change of f(x) = −x³ − 3 on [0, 6].
A: (f(6) − f(0))/(6 − 0) = (−219 − (−3))/6 = −216/6 = −36.
Limits are the gateway to derivatives. The derivative is defined as lim (h→0) (f(x + h) − f(x))/h, so everything in this section underpins the differentiation rules you will use in the rest of the course. Continuity matters again when you study the Intermediate Value Theorem (IVT) and later the Fundamental Theorem of Calculus, both of which require the function to be continuous on a closed interval. Asymptote analysis returns when you sketch curves and find long-run behaviour of functions in the curve-sketching unit.
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