Source: Calc AB Unit 1 Lecture Notes, Advanced Calculus I (UCF)
Difficulty: Introductory. No calculus prerequisites, but a solid grasp of algebra (factoring, rationalising, basic trig identities) is assumed.
Tags: limits, one-sided limits, two-sided limits, limit does not exist, DNE, finding limits analytically, rationalising, special trig limits, sinx/x, limits at infinity, infinite limits, horizontal asymptotes, vertical asymptotes, end behaviour, Calc AB Unit 1
Limits are the foundation of everything in calculus. Before you can define a derivative or an integral, you need a precise way to talk about what a function is "approaching." This unit introduces the language and mechanics of limits, including the cases where they exist, the cases where they do not, and the algebraic techniques you will use over and over to evaluate them. If you are joining the course late, start here; every later topic depends on this material.
A limit describes the value a function approaches as x gets close to a particular point (or heads towards infinity). You can evaluate limits by direct substitution, algebraic simplification, or by recognising special forms. When the left-hand and right-hand behaviour disagree, or the function blows up or oscillates, the limit does not exist.
Limit (two-sided)
The value a function f(x) approaches as x approaches a given value from both the left and the right simultaneously. Written lim (x→c) f(x) = L. The function does not need to equal L at x = c; it only needs to approach L from both sides.
Think of it as: the number the y-values are "heading towards" as you slide along the curve from either direction.
One-sided limit
The value a function approaches when x comes from only one direction. lim (x→c⁻) is the left-hand limit (x approaches c from values less than c) and lim (x→c⁺) is the right-hand limit (x approaches c from values greater than c).
In simple terms, you are only watching the function's behaviour on one side of the point, not both.
Limit does not exist (DNE)
A limit fails to exist in three classic situations:
The left-hand and right-hand limits are different values.
The function grows without bound (heads to +∞ or −∞).
The function oscillates wildly and never settles on a single value.
Think of it as: the function cannot make up its mind about a single number to approach.
Limits at infinity
The behaviour of f(x) as x itself heads to +∞ or −∞. These limits describe the function's end behaviour and are associated with horizontal asymptotes.
In simple terms, this answers: "What happens to y when x gets unimaginably large (or unimaginably negative)?"
Infinite limits
Limits evaluated at a specific x-value where the function's output grows without bound (positive or negative infinity). These are associated with vertical asymptotes.
In simple terms, the x-value is fixed, but the y-value is blowing up.
Horizontal asymptote
A horizontal line y = L that the graph of f(x) approaches as x → +∞ or x → −∞. Found by evaluating limits at infinity.
Vertical asymptote
A vertical line x = c where f(x) → +∞ or f(x) → −∞. Found by evaluating infinite limits at that x-value.
There is a standard sequence of techniques. Work through them in order:
Direct substitution: Plug the given x-value into the function. If you get a real number with no division by zero or other indeterminate form, you are done.
Algebraic simplification: If substitution produces 0/0 or another indeterminate form, simplify the expression first.
Factor and cancel: Factor numerator and denominator, then cancel the common factor that caused the 0/0.
Rationalise numerators or denominators: When square roots appear, multiply by the conjugate to clear the radical and reveal a factor you can cancel.
Find lim (x→9) of (x − 9) / (√x − 3).
Direct substitution gives 0/0, so simplify.
Multiply numerator and denominator by the conjugate (√x + 3):
$$\frac{x - 9}{\sqrt{x} - 3} \cdot \frac{\sqrt{x} + 3}{\sqrt{x} + 3} = \frac{(x - 9)(\sqrt{x} + 3)}{x - 9} = \sqrt{x} + 3$$
Now substitute x = 9: √9 + 3 = 3 + 3 = 6.
These come up constantly. Memorise them:
lim (x→0) sin x / x = 1
lim (x→0) (1 − cos x) / x = 0
Find lim (x→0) sin 3x / x.
Rewrite to match the special form: sin 3x / x = 3 · (sin 3x / 3x).
As x→0, 3x→0 as well, so sin 3x / 3x → 1.
Result: 3 · 1 = 3.
For rational functions (polynomial divided by polynomial) as x → ±∞:
Top-heavy (degree of numerator > degree of denominator): the limit is ±∞. No horizontal asymptote.
Bottom-heavy (degree of numerator < degree of denominator): the limit is 0. Horizontal asymptote at y = 0.
Equal degree: the limit is the ratio of the leading coefficients.
Result | Statement |
|---|---|
Special trig limit 1 | lim (x→0) sin x / x = 1 |
Special trig limit 2 | lim (x→0) (1 − cos x) / x = 0 |
Top-heavy rational | lim (x→±∞) = ±∞ |
Bottom-heavy rational | lim (x→±∞) = 0 |
Equal-degree rational | lim (x→±∞) = leading coefficient of numerator / leading coefficient of denominator |
Limits describe what happens "in the long run" or "at the boundary." Engineers use limits at infinity to model steady-state behaviour in circuits and systems. The special trig limits appear throughout signal processing and wave analysis whenever small-angle approximations are needed.
"The limit equals f(c)." Not necessarily. The limit depends on the values near c, not the value at c. A function can have a hole at x = c and still have a perfectly well-defined limit there.
"If the function is undefined at a point, the limit doesn't exist." Incorrect. The limit can exist even when f(c) is undefined. The classic example is (x² − 1)/(x − 1) at x = 1.
"sin x / x equals 1 everywhere." It equals 1 only in the limit as x → 0. For any other x, sin x / x is just a number less than 1 (and it varies).
"Top-heavy means the limit is always +∞." The sign depends on the leading coefficients and the direction (x → +∞ vs. x → −∞). It could be −∞.
⚠️ The two special trig limits (sin x / x and (1 − cos x) / x) are tested frequently, often in disguised forms like sin 5x / x or sin 2x / sin 3x.
⚠️ Expect at least one problem where direct substitution gives 0/0 and you must factor or rationalise before evaluating.
⚠️ Know the three reasons a limit can fail to exist; exam questions may ask you to identify which reason applies from a graph.
⚠️ For limits at infinity, the top-heavy/bottom-heavy/equal-degree classification is a quick-and-reliable shortcut. Be ready to state horizontal asymptotes.
True or False: lim (x→0) sin x / x = 0.
Fill in the blank: If the degree of the numerator is less than the degree of the denominator, the limit at infinity is ____.
True or False: A limit can exist at a point where the function is not defined.
Fill in the blank: A vertical asymptote is associated with an ____ limit.
True or False: If the left-hand limit and right-hand limit at a point are different, the two-sided limit still exists.
Answers: 1. False (it equals 1). 2. 0. 3. True. 4. Infinite. 5. False.
Q: Evaluate lim (x→9) (x − 9) / (√x − 3).
A: Rationalise by multiplying by (√x + 3)/(√x + 3). The expression simplifies to √x + 3. Substituting x = 9 gives 6.
Q: Evaluate lim (x→0) sin 5x / x.
A: Rewrite as 5 · (sin 5x / 5x). Since sin u / u → 1 as u → 0, the answer is 5.
Q: State the three conditions under which a limit fails to exist.
A: (1) Left-hand and right-hand limits are not equal. (2) The function grows without bound (unbounded behaviour). (3) The function oscillates without settling on a value.
Q: A rational function has a numerator of degree 3 and a denominator of degree 3 with leading coefficients 4 and 2, respectively. What is the limit as x → ∞?
A: Equal degree, so the limit is the ratio of leading coefficients: 4/2 = 2.
Q: True or false: lim (x→0) (1 − cos x) / x = 1.
A: False. The limit is 0.
This material feeds directly into the definition of the derivative, which is itself a limit (the difference quotient). The continuity conditions covered in Part 2 of this unit also rely on limits. If you go on to study series, you will use limits at infinity to test for convergence.
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