Source: MTH 132 Chapter 1, Michigan State University
Difficulty: Introductory to Intermediate | Prerequisites: Part 1 study notes (Sections 4 and 5), comfort with factoring quadratics, rationalising denominators (conjugate multiplication), absolute value, and basic trig identities.
Section 6 gives you the algebraic toolkit for computing limits without a graph or a table: plug in (direct substitution), handle the cases where that fails (0/0 and N/0), and deploy the Squeeze Theorem when algebra alone is not enough. Section 8 ties limits back to the behaviour of functions: a function is continuous precisely when the limit equals the function value, and the Intermediate Value Theorem tells you what continuous functions guarantee. Together these sections form the bridge between the intuitive limit idea (Section 5) and the derivative (Chapter 2).
To evaluate a limit, start by plugging in. If that gives a number, you are done (direct substitution). If you get 0/0, factor, rationalise, or simplify until you can cancel and try again. If you get N/0, the limit is ±∞ (vertical asymptote). The Squeeze Theorem handles the rare cases where none of those tricks work. A function is continuous at a point when the limit there equals the function value; discontinuities come in three flavours (removable, jump, infinite); and the Intermediate Value Theorem guarantees that a continuous function on [a, b] hits every value between f(a) and f(b).
Limit laws
A set of rules that let you break a complicated limit into simpler pieces: the limit of a sum is the sum of the limits, the limit of a product is the product of the limits, and so on, provided each individual limit exists. In simple terms, you can pull limits apart the same way you pull apart arithmetic.
Direct substitution property
If f is a polynomial, rational function, or root function and a is in the domain of f, then lim (x→a) f(x) = f(a). You just plug in. In simple terms, for "nice" functions, the limit is whatever you get when you substitute.
Indeterminate form 0/0
When both the numerator and denominator approach zero, direct substitution gives 0/0, which tells you nothing about the limit's value. You must simplify (factor, rationalise, find a common denominator) before trying again. Think of it as the function saying "I need more work before I'll give you an answer."
Cancellation theorem
If f(x) = g(x) for all x ≠ a, then lim (x→a) f(x) = lim (x→a) g(x), provided the latter limit exists. This justifies cancelling common factors in 0/0 limits.
Conjugate multiplication
A technique for 0/0 limits involving square roots. Multiply numerator and denominator by the conjugate (change the sign between the terms with the root) to eliminate the root and reveal a cancellable factor.
Squeeze Theorem (Sandwich Theorem)
If f(x) ≤ g(x) ≤ h(x) near a (except possibly at a) and lim (x→a) f(x) = lim (x→a) h(x) = L, then lim (x→a) g(x) = L. Think of it as trapping the mystery function between two known functions that both converge to the same value.
Continuity at a point
A function f is continuous at x = a if and only if lim (x→a) f(x) = f(a). This requires three things: f(a) is defined, the limit exists, and they are equal. In simple terms, you can draw the function through that point without lifting your pen.
Removable discontinuity
A discontinuity that can be "fixed" by redefining f at a single point. Typically arises in the 0/0 case. The limit exists, but either f(a) is undefined or f(a) does not match the limit. Think of it as a hole in the graph that you could fill in with one dot.
Jump discontinuity
A discontinuity where the left-hand and right-hand limits both exist but are not equal. The function "jumps" from one value to another. Typically found in piecewise-defined functions.
Infinite discontinuity
A discontinuity where at least one one-sided limit is ±∞. This is another name for a vertical asymptote.
Intermediate Value Theorem (IVT)
If f is continuous on [a, b] and N is any number between f(a) and f(b) (with f(a) ≠ f(b)), then there exists at least one c in (a, b) such that f(c) = N. In simple terms, a continuous function cannot skip over a value. If it starts below a target and ends above it, it must hit that target somewhere in between.
The first step for any limit is to substitute x = a into the expression. The result tells you which case you are in:
Case 1 (N/N or 0/N): You get a well-defined number. The limit equals that number. Done.
Case 2 (N/0): Numerator is non-zero, denominator is zero. This signals a vertical asymptote. Determine the sign on each side to decide whether the limit is +∞ or −∞.
Case 3 (0/0): Both numerator and denominator are zero. This is the indeterminate form. You must do algebraic work before re-evaluating.
The limit laws guarantee that for polynomials, rational functions (where the denominator is non-zero at a), and root functions (where the radicand is non-negative at a), you can simply plug in.
Example: lim (x→0) (2x² − x)/(3 + x) = (0 − 0)/(3 + 0) = 0/3 = 0.
If you know lim (x→a) f(x) and lim (x→a) g(x), the limit laws let you combine them: lim [5f(x) − xg(x)] = 5·lim f(x) − a·lim g(x).
When substitution gives a non-zero numerator over zero, the function blows up. Determine the sign:
Note the sign of the numerator at a.
Test a value just to the left of a and just to the right to find the sign of the denominator on each side.
Combine: same sign → +∞; opposite signs → −∞.
Example: lim (x→1⁺) (−4)/(1 − x). Numerator is −4 (negative). Denominator: when x is just above 1, (1 − x) is slightly negative. Negative/negative = positive → +∞.
When substitution gives 0/0, the limit might still exist but you need to simplify first. The three main techniques:
Factoring and cancelling
Factor both numerator and denominator.
Cancel the common factor that is causing the 0/0.
Substitute again.
Example: lim (x→1) (x² + 2x − 3)/(x² − 1) = lim (x→1) (x + 3)(x − 1)/((x + 1)(x − 1)) = lim (x→1) (x + 3)/(x + 1) = 4/2 = 2.
Common denominators (for compound fractions)
Combine fractions in the numerator or denominator into a single fraction.
Simplify and cancel.
Example: lim (x→2) [(1/(x − 1)) − 1]/(x − 2). Combine the numerator: (1 − (x − 1))/(x − 1) = (2 − x)/(x − 1). Then the whole expression becomes (2 − x)/((x − 1)(x − 2)) = −1/(x − 1). Substituting x = 2 gives −1.
Conjugate multiplication (for expressions with roots)
Multiply numerator and denominator by the conjugate of the expression containing the root.
The root disappears via the difference-of-squares pattern.
Cancel and substitute.
Example: lim (h→0) (√(9 + h) − 3)/h. Multiply top and bottom by (√(9 + h) + 3). Numerator becomes (9 + h) − 9 = h. Cancel h. Result: 1/(√(9 + h) + 3) → 1/6.
When algebraic simplification cannot resolve a 0/0 limit (often involving trigonometric or oscillating functions), the Squeeze Theorem provides an alternative.
To use it:
Find two simpler functions f(x) and h(x) such that f(x) ≤ g(x) ≤ h(x) near a.
Show that lim (x→a) f(x) = lim (x→a) h(x) = L.
Conclude that lim (x→a) g(x) = L.
Example: lim (x→0) x² cos(2/x). Since −1 ≤ cos(2/x) ≤ 1, we have −x² ≤ x² cos(2/x) ≤ x². Both −x² and x² approach 0 as x→0, so by the Squeeze Theorem, the limit is 0.
When a limit involves |expression|, split into left and right cases:
For x > a (or approaching from the right), |x − a| = x − a.
For x < a (or approaching from the left), |x − a| = −(x − a) = a − x.
Evaluate each one-sided limit separately, then compare.
A function f is continuous at x = a if and only if:
lim (x→a) f(x) = f(a)
This single equation packs in three requirements:
f(a) must be defined.
lim (x→a) f(x) must exist (both one-sided limits exist and agree).
The limit must equal the function value.
If any one of these fails, the function is discontinuous at a.
A function is continuous on an interval if it is continuous at every point in that interval.
Polynomials, rational functions, root functions, and trigonometric functions are all continuous everywhere in their domains. This means you only need to worry about continuity at points that lie outside the domain, or at boundaries of piecewise definitions.
Combinations of continuous functions (sums, differences, products, scalar multiples, quotients where the denominator is non-zero) are also continuous. Compositions of continuous functions are continuous: if g is continuous at a and f is continuous at g(a), then f(g(x)) is continuous at a.
Division by zero: find values where the denominator equals zero.
Square roots of negatives: find values where the radicand is negative.
Boundaries of piecewise-defined functions: check whether the left-hand and right-hand pieces agree at each boundary, and whether the function value matches.
Removable discontinuity
The limit exists at a, but either f(a) is not defined or f(a) does not equal the limit. You can "remove" the discontinuity by redefining f(a) to equal the limit. This typically corresponds to the 0/0 case in limit evaluation.
Example: f(x) = (x² + 5x + 6)/(x² − 3x − 10). Factor: (x + 2)(x + 3)/((x − 5)(x + 2)). At x = −2, both numerator and denominator are zero. Cancel (x + 2) to get (x + 3)/(x − 5). The limit as x→ −2 is 1/(−7) = −1/7. This is a removable discontinuity (a hole).
Infinite discontinuity
At least one one-sided limit is ±∞. This is a vertical asymptote. In the example above, x = 5 gives 8/0, so x = 5 is an infinite discontinuity.
Jump discontinuity
Both one-sided limits exist but are not equal. The function jumps from one value to another. Typically found at boundaries of piecewise functions.
Example: f(x) = { x + 1 if x < 0, 2cos(πx) if 0 ≤ x ≤ 2, 6 − x² if x > 2 }. At x = 0: left-hand limit = 1, right-hand limit = 2cos(0) = 2. Since 1 ≠ 2, there is a jump discontinuity. At x = 2: left-hand limit = 2cos(2π) = 2, right-hand limit = 6 − 4 = 2. Both agree and f(2) = 2, so the function is continuous at x = 2.
To "remove" a removable discontinuity at x = a:
Simplify f(x) algebraically (factor and cancel).
The simplified form g(x) agrees with f(x) for all x ≠ a.
Define g(a) = lim (x→a) f(x).
g is now continuous at a.
Example: f(x) = (3x² − 3x)/(x² − 1) has a removable discontinuity at x = 1. Factor: 3x(x − 1)/((x − 1)(x + 1)) = 3x/(x + 1) for x ≠ 1. Define g(x) = 3x/(x + 1). Then g(1) = 3/2, and g is continuous at 1.
A common exam problem: given f(x) = { cx² + 2x if x < 2, x³ − cx if x ≥ 2 }, find c so that f is continuous everywhere.
At x = 2, set the left-hand limit equal to the right-hand limit: c(4) + 4 = 8 − 2c. Solve: 4c + 4 = 8 − 2c → 6c = 4 → c = 2/3.
If f is continuous on [a, b] and N is any number between f(a) and f(b), then there exists at least one c in (a, b) with f(c) = N.
The IVT does not tell you what c is. It only guarantees that c exists.
The most common use: proving that an equation has a solution (a root) in an interval.
IVT root-finding procedure:
Rearrange the equation so one side is zero: f(x) = 0.
Confirm f is continuous on [a, b].
Compute f(a) and f(b). If one is positive and the other is negative, then by the IVT there is a root in (a, b).
Example: Show x⁴ + x − 3 = 0 has a root in (1, 2). f(1) = 1 + 1 − 3 = −1 < 0. f(2) = 16 + 2 − 3 = 15 > 0. Since f is continuous (polynomial) and changes sign, the IVT guarantees a root in (1, 2).
Example: Prove cos(x) = x³ has a solution. Let g(x) = cos(x) − x³. g(0) = 1 > 0. g(1) = cos(1) − 1 ≈ −0.46 < 0. g is continuous, so the IVT guarantees a root in (0, 1).
Limit laws (assuming lim f and lim g both exist)
lim [f(x) + g(x)] = lim f(x) + lim g(x)
lim [c · f(x)] = c · lim f(x)
lim [f(x) · g(x)] = lim f(x) · lim g(x)
lim [f(x)/g(x)] = lim f(x) / lim g(x), provided lim g(x) ≠ 0
lim c = c
lim x = a
lim xⁿ = aⁿ
lim [f(x)]ⁿ = [lim f(x)]ⁿ
Direct substitution property
For polynomials, rational functions, and root functions where a is in the domain: lim (x→a) f(x) = f(a).
Squeeze Theorem
If f(x) ≤ g(x) ≤ h(x) near a and lim f(x) = lim h(x) = L, then lim g(x) = L.
Continuity condition
f is continuous at a ⟺ lim (x→a) f(x) = f(a).
Intermediate Value Theorem
f continuous on [a, b], f(a) ≠ f(b), N between f(a) and f(b) ⟹ there exists c ∈ (a, b) with f(c) = N.
Continuity underpins most physical reasoning. Temperature varies continuously with position, so if it is 15°C on the ground floor and 22°C on the top floor, every temperature between 15 and 22 must occur somewhere in the building (IVT). The same logic applies to elevation on a hiking trail, pressure in a pipeline, or concentration in a chemical reaction. Engineers rely on continuity when modelling any system where quantities change smoothly rather than jumping.
Students often assume that if they get 0/0, the limit does not exist. In fact, 0/0 is an indeterminate form, meaning the limit could be any number (or could fail to exist). You must simplify further.
Students sometimes cancel factors without checking whether the cancellation introduces a restriction. Remember: cancelling (x − a) is valid only for x ≠ a, but that is fine for limits because limits do not care about the value at a.
Students confuse "f is discontinuous at a" with "f is not defined at a." A function can be defined at a and still be discontinuous there (e.g. a piecewise function whose pieces do not match at the boundary).
Students misapply the IVT by forgetting to verify continuity. The IVT requires continuity on the entire closed interval. If f has a discontinuity in [a, b], the theorem does not apply.
⚠️ The "plug in first" triage (N/N, N/0, or 0/0) is the single most important procedural skill in this chapter. Every limit problem starts there.
⚠️ Factoring-and-cancelling 0/0 limits are the most common exam question type for Section 6. Practise until the factoring step is automatic.
⚠️ Conjugate multiplication problems (roots in a 0/0 limit) appear on nearly every exam. Know the pattern: multiply by the conjugate, simplify, cancel, substitute.
⚠️ Piecewise continuity problems ("find c so that f is continuous") are a staple. Set left-hand limit = right-hand limit at the boundary and solve for c.
⚠️ IVT problems on the exam follow a strict template: state that f is continuous on [a, b], compute f(a) and f(b), note the sign change, cite the IVT. Do not skip any step.
⚠️ Classifying discontinuities (removable, jump, infinite) from a formula is frequently tested. Factor the rational function first; the 0/0 points are removable, the N/0 points are infinite, and piecewise boundaries with mismatched limits are jumps.
Plugging x = 3 into (x² − 9)/(x − 3) gives 0/0. The limit as x→3 is __________.
True or False: If f is continuous at x = 5, then lim (x→5) f(x) = f(5).
The three types of discontinuity covered in this course are __________, __________, and __________.
True or False: The IVT can be used to find the exact value of a root.
To evaluate lim (h→0) (√(4 + h) − 2)/h, you should multiply by the __________ of the numerator.
Answers: 1. 6 (factor and cancel: (x + 3)(x − 3)/(x − 3) = x + 3 → 6). 2. True (that is the definition of continuity). 3. Removable, jump, infinite. 4. False (the IVT guarantees existence of a root in an interval but does not give its exact value). 5. Conjugate, i.e. (√(4 + h) + 2).
Q: Evaluate lim (x→9) (√x − 3)/(9 − x).
A: Substitution gives 0/0. Multiply numerator and denominator by (√x + 3): numerator becomes x − 9, denominator becomes (9 − x)(√x + 3) = −(x − 9)(√x + 3). Cancel (x − 9): result is −1/(√x + 3). Substitute x = 9: −1/(3 + 3) = −1/6.
Q: Evaluate lim (t→1) |t − 1|/(1 − t).
A: From the right (t→1⁺): |t − 1| = t − 1, so (t − 1)/(1 − t) = −1. From the left (t→1⁻): |t − 1| = 1 − t, so (1 − t)/(1 − t) = 1. Left limit ≠ right limit, so the two-sided limit does not exist.
Q: Suppose 2x + 1 ≤ g(x) ≤ x² − 2x + 5 for all x. Find lim (x→2) g(x).
A: lim (x→2) (2x + 1) = 5. lim (x→2) (x² − 2x + 5) = 4 − 4 + 5 = 5. Both bounds approach 5, so by the Squeeze Theorem, lim (x→2) g(x) = 5.
Q: Determine and classify the discontinuities of f(x) = (x² + 5x + 6)/(x² − 3x − 10).
A: Factor: (x + 2)(x + 3)/((x − 5)(x + 2)). Denominator is zero at x = 5 and x = −2. At x = −2: both numerator and denominator are zero (0/0). Cancel (x + 2) to get (x + 3)/(x − 5). Limit is 1/(−7) = −1/7. Removable discontinuity. At x = 5: numerator is (5 + 3) = 8 ≠ 0, denominator is 0 (N/0). Infinite discontinuity (vertical asymptote).
Q: Suppose f is continuous everywhere with f(1) = 5, f(3) = 2, f(11) = −1. Which must be true: (A) f(c) = 0 for some c ∈ [1, 3]; (B) f(c) = 0 for some c ∈ [−1, 5]; (C) f(a) = f(b) for some a ≠ b; (D) f(c) = 4 for some c ∈ [1, 3]; (E) None of these?
A: (D). Since f(1) = 5 and f(3) = 2, and 4 is between 2 and 5, the IVT guarantees f(c) = 4 for some c ∈ (1, 3). (A) is not guaranteed because 0 is not between 2 and 5. (B) involves x = −1 where we have no data. (C) is true (it follows from the IVT applied twice) but is not listed as the standard answer on this exam. The intended answer is (D).
Q: Use the IVT to show that ³√x + x = 7 has a solution.
A: Let f(x) = ³√x + x − 7. f is continuous everywhere (sum of a root function and a polynomial). f(0) = 0 + 0 − 7 = −7 < 0. f(8) = 2 + 8 − 7 = 3 > 0. Since f changes sign on [0, 8], the IVT guarantees a root c ∈ (0, 8) where f(c) = 0, i.e. ³√c + c = 7.
The limit laws from Section 6 are the machinery behind every derivative you will compute in Chapter 2. The 0/0 case, in particular, is precisely the form that the definition of the derivative takes: lim (h→0) [f(a + h) − f(a)]/h is always 0/0, and the factoring or rationalising techniques you practise here are the same ones you will use to differentiate from the definition. Continuity is a prerequisite for nearly every major theorem in calculus, including the Mean Value Theorem (Chapter 3) and the Fundamental Theorem of Calculus (Chapter 5). The IVT connects to root-finding algorithms like the Bisection Method in numerical analysis.
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