Difficulty: Introductory | Prerequisites: Basic algebra (factoring quadratics, simplifying rational expressions)
Limits are the foundation of everything in calculus: derivatives, integrals, and continuity all rest on this one idea. This is the first concept you will meet in the course, and it comes back in every chapter that follows. You need to be comfortable with factoring polynomials and simplifying fractions before diving in. If you missed the algebra review, go back to that first.
A limit describes what value a function approaches as x gets close to some number, without x needing to reach that number. You can often find a limit by plugging in directly, but when that gives 0/0 you need to factor and simplify first. One-sided limits look at the function from the left or right only, and the full (two-sided) limit exists only when both sides agree.
Limit (L)
The value that f(x) approaches as x gets arbitrarily close to a number c, without x ever equalling c. Written as lim(x→c) f(x) = L.
In simple terms, it is the answer to: "Where is this function heading as I get close to c?"
One-sided limit (left-hand)
The value f(x) approaches as x comes toward c from values less than c. Notated with a superscript minus: lim(x→c⁻) f(x).
Think of it as sneaking up on c from the left on a number line.
One-sided limit (right-hand)
The value f(x) approaches as x comes toward c from values greater than c. Notated with a superscript plus: lim(x→c⁺) f(x).
Think of it as sneaking up on c from the right.
Two-sided limit
The ordinary limit lim(x→c) f(x). It exists only when both one-sided limits exist and are equal to each other.
In simple terms, both directions have to agree, or there is no limit.
Indeterminate form (0/0)
The result you get when direct substitution gives 0 in both the numerator and denominator. It does not mean the limit is zero or undefined; it means more work is needed (usually factoring).
Think of it as a "try again" signal, not an answer.
Direct substitution
The simplest limit technique: plug c straight into f(x). Works whenever the function is defined and continuous at c.
In simple terms, just put the number in and see what you get.
Factoring technique
When direct substitution gives 0/0, factor the numerator and denominator, cancel the common factor, then substitute again.
Think of it as clearing the roadblock so you can substitute cleanly.
The limit of f(x) as x approaches c describes the behaviour of the function near c, not at c. The function does not need to be defined at c for the limit to exist. What matters is the trend: as x values get closer and closer to c (from both sides), what single value does f(x) settle toward?
If f(x) settles toward one value L, then lim(x→c) f(x) = L.
If f(x) heads toward different values from the left and right, the limit does not exist (DNE).
When you know lim(x→c) f(x) = s and lim(x→c) g(x) = t, you can combine them:
Product rule for limits: lim(x→c) [f(x) · g(x)] = s · t
Sum rule: lim(x→c) [f(x) + g(x)] = s + t
Quotient rule: lim(x→c) [f(x) / g(x)] = s / t, provided t ≠ 0
Constant multiple: lim(x→c) [k · f(x)] = k · s
Power rule: lim(x→c) [f(x)]ⁿ = sⁿ
These rules let you break complicated limits into simpler pieces. You evaluate each piece, then combine the results.
A left-hand limit (x→c⁻) looks at what happens as x approaches c from values smaller than c. The minus sign in the superscript means "from the left."
A right-hand limit (x→c⁺) looks at what happens as x approaches c from values larger than c. The plus sign means "from the right."
The two-sided limit lim(x→c) f(x) exists if and only if:
The left-hand limit exists, and
The right-hand limit exists, and
Both are equal
Example from Section 2.1, Q6: For the piecewise function f(x) = (−x²−4)/4 when x < 2 and f(x) = x²−3x when x ≥ 2:
lim(x→2⁻) f(x) = (−4−4)/4 = −8/4 = −2
lim(x→2⁺) f(x) = 4−6 = −2
Both sides equal −2, so lim(x→2) f(x) = −2. The limit exists.
Step 1, always try direct substitution first. Plug c into the function. If you get a real number, that is your answer.
Example (Q4): lim(x→−5) √(7−4x)
Substitute: √(7−4(−5)) = √(7+20) = √27 = 3√3
Direct substitution worked. Done.
Step 2, if you get 0/0, factor and cancel. The 0/0 result is called an indeterminate form. It means the limit likely exists but you need to simplify.
Example (Q5): lim(x→2) (x²−4)/(x²−3x+2)
Direct substitution gives (4−4)/(4−6+2) = 0/0. Indeterminate.
Factor numerator: x²−4 = (x−2)(x+2)
Factor denominator: x²−3x+2 = (x−1)(x−2)
Cancel (x−2): (x+2)/(x−1)
Substitute x = 2: (2+2)/(2−1) = 4/1 = 4
The limit is 4.
A piecewise function uses different formulas for different ranges of x. To find its limit at the boundary:
Use the left-hand piece for lim(x→c⁻)
Use the right-hand piece for lim(x→c⁺)
Compare the two results
If they match, the two-sided limit equals that shared value. If they differ, the limit does not exist.
This is one of the most common setups on exams, because it tests whether you understand one-sided limits, piecewise evaluation, and the existence condition all at once.
When given a graph of f(x), you find a limit by looking at where the curve is heading, not where the dot sits.
To find lim(x→c) f(x): trace the curve from both sides toward x = c. If both sides approach the same y-value, that y-value is the limit.
An open circle at a point means the function is not defined there, but the limit can still exist.
A filled circle at a different height from the curve means f(c) exists but differs from the limit.
DNE (Does Not Exist): if the curve approaches different y-values from the left and right, the limit does not exist.
Example (Q7): From the graph in the warmup:
lim(x→0) f(x) = 1 (the curve approaches y = 1 from both sides)
lim(x→1) f(x) = 2 (the curve approaches y = 2 from both sides)
lim(x→2) f(x) = DNE (the curve approaches different values from left and right)
\lim_{x \to c} f(x) = L\lim_{x \to c} [f(x) \cdot g(x)] = \left[\lim_{x \to c} f(x)\right] \cdot \left[\lim_{x \to c} g(x)\right]\lim_{x \to c} \frac{f(x)}{g(x)} = \frac{\lim_{x \to c} f(x)}{\lim_{x \to c} g(x)}, \quad \text{provided } \lim_{x \to c} g(x) \neq 0Key factoring pattern used in this section:
x² − 4 = (x−2)(x+2), a difference of squares
x² − 3x + 2 = (x−1)(x−2), a standard quadratic factorisation
In business calculus, limits underpin marginal analysis: marginal cost and marginal revenue are defined as limits of difference quotients. When an economist says "the cost of producing one more unit," they are describing a limit. The same idea appears in finance when computing instantaneous rates of return on investments.
Students often think the limit equals the function's value at c. It does not. The limit is about the approach, and f(c) can differ from the limit or not exist at all.
Students often think 0/0 means the limit is zero, or that it means the limit is undefined. Neither is correct. 0/0 is an indeterminate form that requires further work.
Students forget to check both sides for piecewise functions. Evaluating just one piece and calling it the limit will cost marks if the other side gives a different value.
Students mix up the notation: x→c⁻ means from the left (values less than c), not that the limit is negative.
⚠️ Expect at least one problem where direct substitution gives 0/0 and you must factor. This is the most commonly tested technique in Section 2.1.
⚠️ Piecewise limit problems (checking both one-sided limits) appear regularly on exams and quizzes.
⚠️ Know the difference between "the limit exists" and "the function is defined." You may be asked to explain why a limit exists even when f(c) is undefined.
⚠️ Graph-reading limit questions are common on exams. Be comfortable identifying limits, open/closed circles, and DNE from a picture.
True or False: The limit of f(x) as x→c requires f(c) to be defined.
False. The limit depends on the behaviour near c, not at c.
True or False: If direct substitution gives 0/0, the limit does not exist.
False. 0/0 is indeterminate, meaning you need to simplify (usually by factoring) and try again.
Fill in the blank: The two-sided limit exists only when the left-hand limit and the right-hand limit are both ______ and ______.
Existent, equal to each other.
True or False: The superscript minus in lim(x→3⁻) means the limit is negative.
False. It means x approaches 3 from the left (from values less than 3).
Fill in the blank: When evaluating a limit of a piecewise function at the boundary, you must check the ______ limit and the ______ limit separately.
Left-hand, right-hand.
Q: Evaluate lim(x→3) (x²−9)/(x−3).
A: Direct substitution gives 0/0. Factor: (x²−9) = (x−3)(x+3). Cancel (x−3). The limit becomes lim(x→3) (x+3) = 6.
Q: Given f(x) = x+1 for x < 4 and f(x) = 2x−3 for x ≥ 4, does lim(x→4) f(x) exist?
A: Left-hand limit: 4+1 = 5. Right-hand limit: 2(4)−3 = 5. Both equal 5, so the limit exists and equals 5.
Q: From a graph, the curve approaches y = 3 from the left of x = 1 and y = 5 from the right of x = 1. What is lim(x→1) f(x)?
A: The limit does not exist (DNE) because the one-sided limits are not equal.
Q: What does lim(x→c⁻) f(x) mean in words?
A: It is the value that f(x) approaches as x gets close to c from values less than c (from the left on a number line).
Q: Evaluate lim(x→−2) √(3−5x).
A: Direct substitution: √(3−5(−2)) = √(3+10) = √13. The limit is √13.
This connects to continuity (Section 2.2), which formally asks: does the limit equal the function value? A function is continuous at c precisely when lim(x→c) f(x) = f(c).
Limits are also the machinery behind derivatives (Chapter 3). The derivative is defined as a limit of a difference quotient, so every derivative you compute later is a limit problem in disguise.
The factoring and simplification skills here carry directly into rational function analysis and finding where functions have holes versus vertical asymptotes.
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