Difficulty: Introductory to Intermediate | Prerequisites: Algebra II / pre-calculus fluency with functions, factoring, and basic trigonometry.
Limits are the foundational idea underneath all of calculus. Every major concept you will meet in this course, from derivatives to integrals, is defined using a limit. This section of MTH 132 asks you to understand what it means for a function to approach a value, how to compute that value using algebraic techniques and limit laws, and how to make the idea rigorous with the epsilon-delta definition. If you are comfortable evaluating limits by factoring, rationalising, and applying the standard trig identities, you have the toolkit for the rest of the course.
A limit describes the value a function approaches as its input gets close to some target. You evaluate limits by direct substitution when possible, or by factoring, rationalising, or using special identities (like sin(x)/x → 1) when substitution gives 0/0. The epsilon-delta definition makes "approaches" precise: for every tolerance ε around the output, you can find a tolerance δ around the input that keeps the function within ε.
Limit
The value that f(x) approaches as x approaches some number c. Written lim(x→c) f(x) = L.
In simple terms, it is the y-value the graph is heading towards, even if the function never lands on it.
Average rate of change
The ratio [f(b) − f(a)] / (b − a) over an interval [a, b]. Geometrically, it is the slope of the secant line through (a, f(a)) and (b, f(b)).
Think of it as: "how fast was the quantity changing on average between two points?"
Instantaneous rate of change
The limit of the average rate of change as the interval shrinks to zero: lim(h→0) [f(a+h) − f(a)] / h. This is the slope of the tangent line at x = a, and it equals the derivative f'(a).
In simple terms, it is the speed at one exact moment, not over a stretch.
One-sided limit
The value f(x) approaches as x comes from one direction only. lim(x→c⁺) means from the right; lim(x→c⁻) means from the left. The two-sided limit exists only when both one-sided limits exist and are equal.
Sum Law
If lim(x→c) f(x) and lim(x→c) g(x) both exist, then lim(x→c) [f(x) + g(x)] = lim(x→c) f(x) + lim(x→c) g(x). The same principle lets you split off a constant term (like + 5) and evaluate it independently.
Product Law / Constant Multiple Law
lim(x→c) [k · f(x)] = k · lim(x→c) f(x). You can pull a constant factor out of a limit.
Quotient Law
lim(x→c) [f(x)/g(x)] = lim(x→c) f(x) / lim(x→c) g(x), provided the denominator limit is not zero.
Root Law
lim(x→c) √[f(x)] = √[lim(x→c) f(x)], provided the limit inside the root is non-negative. If lim(x→3) f(x) = 6, then lim(x→3) √f(x) = √6.
Squeeze Theorem (Sandwich Theorem)
If g(x) ≤ f(x) ≤ h(x) near c, and lim(x→c) g(x) = lim(x→c) h(x) = L, then lim(x→c) f(x) = L.
Think of it as: if two bounding functions pinch together at the same value, whatever is trapped between them must go there too. Used classically to show lim(x→0) x² cos(1/x) = 0.
Epsilon-delta (ε-δ) definition of a limit
lim(x→c) f(x) = L means: for every ε > 0, there exists a δ > 0 such that whenever 0 < |x − c| < δ, we have |f(x) − L| < ε.
In simple terms, you pick any tolerance band around L (that is ε), and I can find a neighbourhood around c (that is δ) so the function stays inside your band.
Indeterminate form 0/0
When direct substitution gives 0/0, the limit may still exist but requires algebraic work (factoring, rationalising, or trig identities) to resolve. This is the most common "case" tested on the MTH 132 midterm.
Formula: [f(b) − f(a)] / (b − a)
Geometrically, this is the slope of the secant line joining two points on the graph.
Example (Michigan population data): average rate of change from 2011 to 2016 = (9,928,300 − 9,876,213) / (2016 − 2011) = 52,087 / 5 = 10,417.4 people per year.
Example (f(x) = √(x − 1) on [3, 8]): f(8) = √7, f(3) = √2, so average rate = (√7 − √2) / 5 ≈ 0.2.
Example (h(t) = 10t − 2t² on [1, 2]): h(1) = 8, h(2) = 12, so average velocity = (12 − 8) / (2 − 1) = 4 m/s.
The limit of the average rate of change as the interval width h → 0.
Visually: the slope of the tangent line to the curve at a single point.
For population data at year 2014, the instantaneous rate of change is the slope of the tangent line to the population curve at the point corresponding to 2014.
For a linear function g(x) = 2x − 3, the instantaneous rate of change is the same everywhere: it equals the slope, which is 2.
If f is continuous at c, then lim(x→c) f(x) = f(c). Just plug in.
This works for polynomials, rational functions (when the denominator is nonzero), and compositions of continuous functions.
When substitution gives 0/0, you must simplify before taking the limit.
Factoring: lim(x→2) (x² − 4)/(x − 2) = lim(x→2) (x+2)(x−2)/(x−2) = lim(x→2) (x+2) = 4.
Factoring (exam-style): lim(x→1) (x² + 7x − 8)/(x² − 1). Both numerator and denominator are 0 at x = 1. Factor: (x+8)(x−1) / (x+1)(x−1) = (x+8)/(x+1). Evaluate: 9/2.
Rationalising: lim(t→0) (√(3+t) − √3)/t. Multiply top and bottom by (√(3+t) + √3): result simplifies to 1/(√(3+t) + √3) → 1/(2√3) as t → 0.
lim(x→0) sin(x)/x = 1. This is the foundational trig limit.
lim(x→0) sin(kx)/(mx) = k/m. Scale both the argument and the denominator.
Example: lim(x→0) sin(6x)/(2x) = 6/2 = 3.
lim(x→0) tan(kx)/(mx) = k/m, because tan(x) = sin(x)/cos(x) and cos(0) = 1.
Example: lim(x→0) tan(2x)/(4x) = 2/4 = 1/2.
lim(x→∞) sin(x)/x = 0. The numerator oscillates between −1 and 1 while the denominator grows without bound, so the fraction is squeezed to 0.
To evaluate lim(x→0) x² cos(1/x): note that −1 ≤ cos(1/x) ≤ 1, so −x² ≤ x² cos(1/x) ≤ x². Since lim(x→0) (−x²) = lim(x→0) x² = 0, the Squeeze Theorem gives the answer as 0.
lim(x→1⁺) 1/(x−1) = +∞. As x approaches 1 from the right, x − 1 is a small positive number, so 1/(x−1) grows without bound.
If the left-hand and right-hand limits at a point are not equal, the two-sided limit does not exist.
For y = x², the slope of the secant line on [1, 1+h] is ((1+h)² − 1)/h = (2h + h²)/h = 2 + h.
As h → 0, the secant slope → 2, which is the slope of the tangent line at x = 1.
In general, as h → 0, the limit of the secant slope becomes the slope of the tangent line (the derivative).
To prove lim(x→1)(2x + 2) = 4 with ε = 0.5:
|f(x) − L| = |2x + 2 − 4| = |2x − 2| = 2|x − 1|.
We need 2|x − 1| < 0.5, so |x − 1| < 0.25.
The largest δ that works is δ = 0.25.
General template for lim(x→c)(mx + b) = mc + b:
|f(x) − L| = |m| · |x − c|.
Choose δ = ε / |m|.
Example: lim(x→3)(2x − 1) = 5. Let ε > 0 be given. Choose δ = ε/2. Then 0 < |x − 3| < δ implies |2x − 1 − 5| = 2|x − 3| < 2(ε/2) = ε.
Average rate of change: [f(b) − f(a)] / (b − a)
Derivative (limit definition): f'(x) = lim(h→0) [f(x+h) − f(x)] / h
Fundamental trig limit: lim(x→0) sin(x)/x = 1
Scaled trig limit: lim(x→0) sin(kx)/(mx) = k/m
Tangent trig limit: lim(x→0) tan(kx)/(mx) = k/m
ε-δ template for linear functions: given lim(x→c)(mx + b) = mc + b, choose δ = ε/|m|
Average rate of change is how you measure growth between two data points: population growth per year, average speed over a journey, revenue change between quarters. Instantaneous rate of change is what a car's speedometer reads at a given moment, or the rate at which a drug concentration in the bloodstream is changing right now.
Students often confuse the average rate of change (secant slope over an interval) with the instantaneous rate of change (tangent slope at a single point). The first uses two points; the second takes a limit.
When a limit gives 0/0, many students write "does not exist." The 0/0 form is indeterminate, meaning more work is needed. The limit often does exist after simplification.
Students sometimes apply l'Hôpital's Rule in MTH 132 when the course expects algebraic methods (factoring, rationalising). Use the techniques taught in class.
In ε-δ proofs, students sometimes pick δ first and then try to find ε. The logic goes the other way: ε is given, and you derive δ from it.
⚠️ The 0/0 factoring case appears repeatedly across multiple choice, short answer, and long answer sections. Be ready to factor both numerator and denominator and cancel the common factor.
⚠️ Know the special trig limits cold: sin(kx)/(mx) = k/m and tan(kx)/(mx) = k/m. These appear as standalone questions.
⚠️ Epsilon-delta proofs are tested in both multiple choice (finding δ given ε for a linear function) and as a full long-answer question. Practise the scratchwork-then-verification format.
⚠️ Average rate of change from a data table is a reliable exam question. Just compute [f(b) − f(a)] / (b − a) carefully.
⚠️ The Squeeze Theorem is tested through x² cos(1/x), which requires recognising the bounding functions −x² and x².
True or False: lim(x→0) sin(5x)/(5x) = 5.
Fill in the blank: The average rate of change of f on [a, b] equals the slope of the ______ line.
True or False: If direct substitution gives 0/0, the limit does not exist.
Fill in the blank: In an ε-δ proof for a linear function f(x) = mx + b, you choose δ = ε / ______.
True or False: lim(x→∞) sin(x)/x = 1.
Answers: 1. False (it equals 1). 2. secant. 3. False (it is indeterminate, not nonexistent). 4. |m|. 5. False (it equals 0).
Q: Using the Michigan population data, what is the average rate of change from 2013 to 2015?
A: (9,917,715 − 9,898,982) / (2015 − 2013) = 18,733 / 2 = 9,366.5 people per year.
Q: Evaluate lim(x→2) (x² − 4)/(x − 2).
A: Factor the numerator as (x+2)(x−2), cancel (x−2), and get lim(x→2)(x+2) = 4.
Q: Evaluate lim(t→0) (√(3+t) − √3)/t.
A: Multiply numerator and denominator by (√(3+t) + √3). The numerator becomes (3+t) − 3 = t. Cancel t, giving 1/(√(3+t) + √3). As t → 0 this becomes 1/(2√3).
Q: Evaluate lim(x→0) sin(6x)/(2x).
A: Rewrite as 3 · [sin(6x)/(6x)]. As x → 0, sin(6x)/(6x) → 1, so the answer is 3.
Q: In an ε-δ proof for lim(x→3)(2x − 1) = 5, what δ should you choose?
A: |2x − 1 − 5| = 2|x − 3| < ε requires |x − 3| < ε/2. Choose δ = ε/2.
Q: For y = x², what is the slope of the secant line on [1, 1+h]?
A: ((1+h)² − 1²)/h = (1 + 2h + h² − 1)/h = (2h + h²)/h = 2 + h.
Q: What is lim(x→0) x² cos(1/x)?
A: By the Squeeze Theorem: −x² ≤ x² cos(1/x) ≤ x², and both bounds go to 0 as x → 0. So the limit is 0.
Limits feed directly into the definition of the derivative (covered in the next unit). Every derivative is a limit of a difference quotient, so the algebraic techniques you practise here (factoring, rationalising) come back when computing derivatives from first principles. The Squeeze Theorem reappears in integration and in sequences/series later in the calculus sequence.
limit, limit laws, sum law, product law, quotient law, root law, Squeeze Theorem, sandwich theorem, epsilon-delta, ε-δ proof, 0/0 indeterminate form, direct substitution, factoring limits, rationalising limits, conjugate method, sin(x)/x, special trig limits, average rate of change, instantaneous rate of change, secant line slope, tangent line slope, difference quotient, MTH 132, Calculus I, Michigan State University