Learning objectives
Write one-sided and two-sided limits in correct notation and explain that a two-sided limit exists only when both one-sided limits exist and agree, whatever value the function takes at that point
Estimate a limit from a graph or a table, evaluate limits with the limit laws, and resolve 0/0 by factoring, rationalizing, clearing a compound fraction or using the standard trigonometric limits
Apply the squeeze theorem, classify a discontinuity as removable, jump or infinite, and choose parameters that make a piecewise function continuous
Connect infinite limits to vertical asymptotes and limits at infinity to horizontal asymptotes, compare growth rates, and apply the intermediate value theorem on a closed interval
Key definitions
Limit
lim f(x) = L as x → c means f(x) can be forced as close to L as required by taking x close enough to c but not equal to c. The value f(c) plays no part.
One-sided limit
A limit computed from one side only: x → c⁻ uses values below c and x → c⁺ uses values above c.
Indeterminate form
A limit whose shape, such as 0/0 or ∞/∞, is compatible with any answer at all; the shape is a signal to rewrite, never a value.
Squeeze theorem
If g(x) ≤ f(x) ≤ h(x) for all x near c apart from possibly c itself, and lim g(x) = lim h(x) = L as x → c, then lim f(x) = L as x → c.
Continuity at a point
f is continuous at c when f(c) is defined, lim f(x) as x → c exists, and those two numbers are equal.
Intermediate value theorem
If f is continuous on the closed interval [a, b] and N lies between f(a) and f(b), then f(c) = N for at least one c in [a, b].
Explanation
Notation and existence. The statement lim f(x) = L as x → c is a claim about the values of f near c, never about f(c) itself. Because "near" has two sides, the two-sided limit is built from two one-sided limits: lim f(x) as x → c⁻ looks only at inputs below c, and lim f(x) as x → c⁺ only at inputs above c. The two-sided limit exists precisely when both one-sided limits exist and are equal, and it then takes their common value. A limit therefore fails to exist in three recognizable ways: the one-sided limits disagree, the function grows without bound, or the function oscillates forever without settling, as sin(1/x) does near 0. Writing lim f(x) = ∞ is a description of unbounded growth, not a statement that a limit exists.
Graphs, tables and the limit laws. On a graph, trace the curve toward x = c from the left and from the right and read the height each branch approaches, ignoring any dot drawn at c. A table such as x = 2.9, 2.99, 2.999 suggests a value but can never confirm one, because finitely many samples say nothing about the points between them; a function can agree with a polynomial at every value you happen to try and still misbehave elsewhere. Once limits of the parts are known, the limit laws combine them: sums, differences, constant multiples, products and powers pass straight through, and so does a quotient provided the limit of the denominator is not zero.
The form 0/0. Substitute first. A non-zero number divided by zero signals unbounded behavior, so examine each side separately for its sign. The form 0/0 is an instruction to rewrite. Factor and cancel: (2x² + 5x − 3)/(x + 3) = (2x − 1)(x + 3)/(x + 3) → 2(−3) − 1 = −7 as x → −3. Rationalize: (x − 25)/(√x − 5) becomes (x − 25)(√x + 5)/(x − 25) = √x + 5 → 10 as x → 25. Clear a compound fraction: (1/(x + 2) − 1/5)/(x − 3) collapses to −1/(5(x + 2)) → −1/25 as x → 3. Use the standard trigonometric limits, valid in radians: (sin x)/x → 1 and (1 − cos x)/x → 0 as x → 0, rescaling so the angle matches the denominator, as in tan(7x)/(2x) = (7/2)·[sin(7x)/(7x)]·(1/cos 7x) → 7/2.
The squeeze theorem. When no algebraic rewrite is available, trap the function. Since −1 ≤ sin(1/x) ≤ 1 for every x ≠ 0, multiplying through by the non-negative quantity x² gives −x² ≤ x² sin(1/x) ≤ x². Both bounds tend to 0 as x → 0, so the squeeze theorem forces x² sin(1/x) → 0 even though sin(1/x) itself has no limit there. A complete answer names the two bounding functions, states that each has the same limit, and only then draws the conclusion; the inequality alone earns nothing.
Continuity and its failures. f is continuous at c when all three of these hold: f(c) exists, the limit exists, and they agree. Polynomial, rational, root, exponential, logarithmic and trigonometric functions are continuous throughout their domains, so the only places worth testing are gaps in the domain and the joins of a piecewise definition. A removable discontinuity has equal one-sided limits that miss the function value, and is repaired by redefining f(c) as the limit; a jump has unequal one-sided limits; an infinite discontinuity has at least one one-sided limit that is unbounded. At a join, set left-hand limit equal to right-hand limit equal to f(c) and solve for the unknown parameter.
Asymptotes, growth rates and the IVT. Cancel common factors first: a surviving zero of the denominator gives a vertical asymptote, while one that cancels leaves a hole. Test each side for its sign rather than guessing. As x → ±∞ the dominant terms decide: equal degrees give the ratio of leading coefficients, a heavier denominator gives 0, and a heavier numerator gives no horizontal asymptote. Across function families the ordering is ln x much slower than xᵖ for any positive p, which is much slower than eˣ, so x⁴e^(−x) → 0 and (ln x)/√x → 0. Finally, the intermediate value theorem turns continuity into existence: state that f is continuous on the closed interval, show the target value lies between the two endpoint values, and conclude that some c in the interval hits it. It never says how many such c exist or where they are.
Formulae
lim x→c f(x) = L ⇔ lim x→c⁻ f(x) = lim x→c⁺ f(x) = L
lim x→0 (sin x)/x = 1; lim x→0 (1 − cos x)/x = 0 (x in radians)
Squeeze: g ≤ f ≤ h near c and lim g = lim h = L ⇒ lim f = L
Continuous at c ⇔ f(c) exists, lim x→c f(x) exists, and the two are equal
Rational end behavior: equal degrees ⇒ ratio of leading coefficients; smaller numerator degree ⇒ 0
Growth as x → ∞: ln x much slower than xᵖ (p positive) much slower than eˣ
IVT: f continuous on [a, b] and N between f(a) and f(b) ⇒ f(c) = N for some c in [a, b]
Worked examples
q: Let f(x) = (x² − 5x + 6)/(x² − 4). Classify every discontinuity of f and state the equations of all asymptotes of its graph.
steps: Factor both parts: f(x) = (x − 2)(x − 3)/((x − 2)(x + 2)).
The factor x − 2 cancels, so for every x apart from 2 the function equals (x − 3)/(x + 2).
At x = 2 the limit is (2 − 3)/(2 + 2) = −1/4 but f(2) is undefined, so there is a removable discontinuity — a hole at the point (2, −1/4).
At x = −2 the numerator tends to −5 while the denominator tends to 0, so f is unbounded: x = −2 is a vertical asymptote. As x → −2⁻ the factor x + 2 is negative, so f(x) → +∞; as x → −2⁺ it is positive, so f(x) → −∞.
The original numerator and denominator both have degree 2, so as x → ±∞ the limit is the ratio of leading coefficients, 1/1 = 1. The line y = 1 is a horizontal asymptote.
In context
A reaction monitored once a second produces a table of concentrations that seems to settle on a value, but only the rate law confirms the limiting concentration — the clearest everyday reason why numerical evidence alone never settles a limit.
Parking charges that jump the instant a new hour begins are a jump discontinuity, while the terminal velocity a falling skydiver approaches but never reaches is a horizontal asymptote of the velocity curve.
Common misconceptions
Students often read the limit off the dot drawn at x = c. In fact the limit ignores the point at c completely, so moving that dot changes nothing.
Students often stop at 0/0 and write "no limit". In fact 0/0 is an instruction to rewrite the expression, not a conclusion about the answer.
Students often write lim f(x) = ∞ and then claim the limit exists. In fact an infinite limit records unbounded growth, and the limit does not exist as a number.
Students often assume a graph can never cross a horizontal asymptote. In fact (sin x)/x crosses y = 0 infinitely often while still tending to 0.
Study tips
Write down the substitution that produced 0/0 before rewriting; it earns the set-up credit and explains the algebra that follows.
For a piecewise continuity question, write all three conditions with actual numbers rather than naming them in words.
Before quoting the intermediate value theorem, say that f is continuous on the closed interval and show the target value lies between the two endpoint values.
Reasoning
Continuity is the hypothesis behind every existence theorem that follows, from the intermediate value theorem to the mean value theorem and, in Unit 6, the fundamental theorem of calculus.
Summary
A two-sided limit exists only when both one-sided limits exist and are equal.
0/0 means rewrite: factor, rationalize, clear the compound fraction, or use sin x/x → 1.
Continuity at c needs f(c), the limit, and their agreement — all three.
Vertical asymptotes come from infinite limits, horizontal asymptotes from limits at infinity.
The IVT needs continuity on a closed interval and delivers existence only.
Check your understanding
Question 1: lim (2x² + 5x − 3)/(x + 3) as x → −3 equals…
1. −7
2. 0
3. It does not exist
Question 2: As x → ∞, which expression grows the fastest?
1. x⁵
2. e^(0.01x)
3. ln(x⁵)
Question 3: The function f(x) = (x − 4)/|x − 4| has which kind of discontinuity at x = 4?
1. An infinite discontinuity
2. A jump discontinuity
3. A removable discontinuity