- Explain a limit intuitively and with the ε–δ definition
- Compute limits with the limit laws, factoring and the remarkable limits
- Recognise indeterminate forms and resolve them
- Classify discontinuities and apply the intermediate value theorem
How does a speedometer know your speed at one particular instant? Speed is distance divided by time, but in a single instant both the distance and the time are zero, and 0/0 means nothing. The way out is to look at shorter and shorter time intervals and ask which number the ratio approaches. This idea, the limit, is the foundation of all of calculus: derivatives, integrals and series are all defined through limits.
The idea of a limit
Take f(x) = (x² − 1)/(x − 1). At x = 1 it is not defined (0/0), but for every x ≠ 1 we can cancel: f(x) = x + 1. Plug in values near 1: f(0.9) = 1.9, f(0.99) = 1.99, f(0.999) = 1.999, f(1.01) = 2.01. The closer x gets to 1, the closer f(x) gets to 2. We write lim (x→1) f(x) = 2. A limit describes the behaviour near a point; the value at the point itself plays no role and may not even exist.
The number L is the limit of f(x) as x → a if for every ε > 0 there is a δ > 0 such that 0 < |x − a| < δ implies |f(x) − L| < ε. In words: we can make f(x) as close to L as we like (closer than any ε) by taking x close enough to a (closer than δ), but not equal to a.
Think of it as a game. An opponent names a tolerance ε for the output, say 0.001. You must answer with a tolerance δ for the input that keeps f(x) within ε of L. If you can win for every ε, however small, the limit is L. Usually δ is found by working backwards from the inequality |f(x) − L| < ε.
Prove that lim (x→3) (2x + 1) = 7 and find δ for ε = 0.01.
Show solutionHide solution
|(2x + 1) − 7| = |2x − 6| = 2|x − 3|.
2|x − 3| < ε ⇔ |x − 3| < ε/2, so we choose δ = ε/2.
Then 0 < |x − 3| < δ gives |f(x) − 7| = 2|x − 3| < 2 · ε/2 = ε. ∎
For ε = 0.01: δ = 0.005, i.e. every x in (2.995, 3.005) gives f(x) in (6.99, 7.01).
One-sided limits and infinity
Sometimes a function behaves differently on the two sides of a. The left-hand limit lim (x→a⁻) f(x) uses only x < a, the right-hand limit lim (x→a⁺) f(x) only x > a. The two-sided limit exists exactly when both one-sided limits exist and are equal. For f(x) = |x|/x the left limit at 0 is −1 and the right limit is 1, so lim (x→0) |x|/x does not exist.
Limits may also involve infinity. lim (x→0⁺) 1/x = +∞ and lim (x→0⁻) 1/x = −∞, so the graph has a vertical asymptote x = 0. lim (x→∞) 1/x = 0 gives a horizontal asymptote y = 0. For a ratio of polynomials as x → ∞, divide the numerator and the denominator by the highest power of x: lim (x→∞) (3x² + 5)/(2x² − x) = lim (3 + 5/x²)/(2 − 1/x) = 3/2.
- Alim (x→a) f(x), a finite number
- Blim (x→a) g(x), a finite number
- ca constant
The limit laws hold only when both limits A and B exist and are finite; otherwise an indeterminate form may appear.
The two remarkable limits
- xthe angle in radians (not degrees!)
The first remarkable limit
Why is it true? For 0 < x < π/2, comparing in the unit circle the areas of a small triangle, a circular sector and a larger triangle gives sin x < x < tan x. Divide by sin x and flip the fractions: cos x < sin x / x < 1. As x → 0, cos x → 1, so the ratio is trapped between two quantities that tend to 1 and must tend to 1 itself (the squeeze theorem). Because sin x / x is an even function, the result also holds for x < 0. Consequences: lim (x→0) tan x / x = 1 and lim (x→0) (1 − cos x)/x² = 1/2.
- eEuler's number, the base of the natural logarithm
- na natural number that grows without bound
The second remarkable limit. Consequences: lim (n→∞) (1 + k/n)ⁿ = eᵏ, lim (x→0) (eˣ − 1)/x = 1, lim (x→0) ln(1 + x)/x = 1.
The second limit has a money story. If 1 manat earns 100% interest per year and the interest is added n times a year, after one year you have (1 + 1/n)ⁿ manat: yearly 2, monthly ≈ 2.613, daily ≈ 2.715 manat. Even with continuous compounding the amount never exceeds e ≈ 2.718 manat.
a) lim (x→2) (x² − 4)/(x² − 5x + 6)
b) lim (x→0) sin 5x / x
c) lim (n→∞) (1 + 2/n)³ⁿ
Show solutionHide solution
The limit is (2 + 2)/(2 − 3) = −4.
b) Multiply and divide by 5: sin 5x / x = 5 · sin 5x / (5x). With t = 5x → 0 the fraction tends to 1, so the limit is 5.
c) (1 + 2/n)³ⁿ = [(1 + 2/n)ⁿ]³ → (e²)³ = e⁶ ≈ 403.4.
Indeterminate forms
If direct substitution produces 0/0, ∞/∞, ∞ − ∞, 0 · ∞, 1^∞, 0⁰ or ∞⁰, the answer is not decided yet: the limit may be any number or may not exist at all. Such an expression is an indeterminate form, and it must be transformed before the limit can be read off.
| Form | Typical method | Example |
|---|---|---|
| 0/0 | factor and cancel, rationalise, equivalents, L'Hôpital's rule | (x² − 1)/(x − 1) → 2 (x → 1) |
| ∞/∞ | divide by the highest power of x | (3x² + 5)/(2x² − x) → 3/2 (x → ∞) |
| ∞ − ∞ | common denominator or multiply by the conjugate | √(x² + x) − x → 1/2 (x → ∞) |
| 0 · ∞ | rewrite as a quotient 0/0 or ∞/∞ | x · ln x → 0 (x → 0⁺) |
| 1^∞ | reduce to the second remarkable limit or take ln | (1 + 2/n)³ⁿ → e⁶ (n → ∞) |
Compute lim (x→∞) (√(x² + x) − x).
Show solutionHide solution
√(x² + x) − x = (x² + x − x²)/(√(x² + x) + x) = x/(√(x² + x) + x).
Divide the numerator and the denominator by x (x > 0): 1/(√(1 + 1/x) + 1).
As x → ∞, 1/x → 0, so the limit is 1/(1 + 1) = 1/2.
Continuity and the intermediate value theorem
f is continuous at a if three conditions hold: f(a) is defined, lim (x→a) f(x) exists, and the two are equal: lim (x→a) f(x) = f(a). Informally, the graph can be drawn through a without lifting the pen.
- Removable discontinuity: the limit exists, but f(a) is undefined or different from it, e.g. (x² − 1)/(x − 1) at x = 1. Defining f(1) = 2 fills the gap.
- Jump discontinuity (first kind): both one-sided limits are finite but different, e.g. |x|/x jumps from −1 to 1 at 0.
- Infinite or essential discontinuity (second kind): at least one one-sided limit is infinite or does not exist, e.g. 1/x at 0, or sin(1/x), which oscillates infinitely often near 0.
All elementary functions (polynomials, rational functions, roots, exponentials, logarithms and trigonometric functions) are continuous at every point of their domain. Sums, products, quotients (where the denominator is not 0) and compositions of continuous functions are continuous too. That is why for a continuous function the limit is found by simple substitution: lim (x→2) (x³ + eˣ⁻²) = 8 + 1 = 9.
- [a, b]a closed interval (endpoints included)
- f(a) · f(b) < 0the function has opposite signs at the endpoints
- ∃ c“there exists c”, a root of f
The intermediate value theorem (Bolzano's theorem). More generally, a continuous function takes every value between f(a) and f(b).
Show that x³ + x − 1 = 0 has a root between 0 and 1, and narrow it down to an interval of length 0.25.
Show solutionHide solution
f(0) = −1 < 0 and f(1) = 1 > 0: the signs differ, so by the theorem there is a root in (0, 1).
Midpoint: f(0.5) = 0.125 + 0.5 − 1 = −0.375 < 0 ⇒ the root is in (0.5, 1).
Midpoint: f(0.75) = 0.421875 + 0.75 − 1 = 0.171875 > 0 ⇒ the root is in (0.5, 0.75).
Continuing in the same way gives x ≈ 0.6823.
Key points
- lim (x→a) f(x) = L: f(x) gets arbitrarily close to L when x is close to, but not equal to, a; the rigorous version uses ε and δ.
- A limit exists exactly when the left and right limits exist and are equal.
- lim (x→0) sin x / x = 1 (radians) and lim (n→∞) (1 + 1/n)ⁿ = e.
- 0/0, ∞/∞, ∞ − ∞, 0 · ∞ and 1^∞ are indeterminate: transform the expression first.
- Continuity at a: lim (x→a) f(x) = f(a); discontinuities are removable, jumps, or of the second kind.
- Intermediate value theorem: a continuous function that changes sign on [a, b] has a root in (a, b).
Check yourself
10 questions. Every correct answer earns XP.