A limit asks where the value of a function heads as x gets as close as you like to some number.
x never has to actually arrive there, and the function does not even have to be defined at that point.
Even at a hole, if the values approaching from both sides agree, the limit definitely exists.
In the first picture, raising the approach level brings the blue point on the left and the orange point on the right toward the hole at x = 1, and both of their function values gather at 2.
What a limit is
At x = 1, (x² − 1)/(x − 1) cannot be computed because the denominator is 0. Yet nearby values all crowd toward 2: x = 0.9 gives 1.9, x = 1.1 gives 2.1, x = 0.99 gives 1.99. The limit of a function asks exactly this, without stepping onto the point itself: where do the surrounding values gather?
🎯 The idea of a limit
①Ask what value f(x) approaches as x gets as close as you like to a
②x never has to reach a; only the values on the way matter
③A limit exists only when the left-hand limit (from the left) and the right-hand limit (from the right) agree
④A limit can exist even if f(a) is not defined
20 %
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f(x) = (x² − 1)/(x − 1), with a hole at x = 1; the values at the two points approaching from either side gather at 2
The green line in the first picture matches y = x + 1, but there is a red hollow circle at x = 1. Raising the approach level t moves the blue point in from the left and the orange point in from the right toward x = 1, and their x and f(x) are written above. At t = 95 % the points sit at x = 0.9 and 1.1, with function values 1.9 and 2.1. In the second picture, shrinking ε narrows the orange band, and the blue band δ = ε/2 narrows with it so that the graph stays inside.
💡 What to look for
①The blue point (left) and the orange point (right) both head for height 2
②f(1) is undefined (hollow circle), yet the limit is clearly 2
③Factoring gives (x + 1)(x − 1)/(x − 1), which equals x + 1 whenever x ≠ 1
Saying it precisely with ε and δ (advanced)
It is tempting to see the red hollow circle and conclude there is no limit, but the circle only marks that the function has no value at x = 1. Both points gather at the same height 2, so the limit is 2. Conversely, having a value f(a) does not make it the limit: if the left- and right-hand limits differ, there is no limit even when f(a) exists. And plugging the target into a 0/0 form breaks the expression, so first look for a common factor to cancel.
1.5
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For f(x) = 2x, the ε band (orange) around L = 4 and the δ band (blue) around a = 2; shrinking ε shrinks δ = ε/2 too
Definition of a limit (ε-δ)
limx→a f(x) = L
For every ε > 0 there is a δ > 0 such that 0 < |x − a| < δ implies |f(x) − L| < ε
🔑 Reading ε and δ
①ε is the error allowed in the y-direction (orange band)
②δ is the width in the x-direction that keeps f(x) inside the ε band (blue band)
③If a suitable δ can always be found however small ε is, the limit exists
④This definition goes beyond the usual high-school course
Basic properties of limits
2
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f(x) = x²/3 (green) and k · f(x) (orange); as x → 3, f(x) approaches 3 and k · f(x) approaches 3k
Limits of sums, products and quotients
lim [f(x) ± g(x)] = lim f(x) ± lim g(x)
When the limits exist, the limit of a sum, difference, product or quotient is computed from them (for a quotient, the limit of the denominator must not be 0)
Constant multiples
lim k·f(x) = k · lim f(x)
A constant can be taken outside the limit
A preview of key limits (calculus)
Basic trigonometric limit
limx→0sin xx = 1
Covered later in calculus; x is measured in radians
Exponential limit
limx→0ex - 1x = 1
Covered later in calculus; e is the base of the natural logarithm
Logarithmic limit
limx→0ln(1 + x)x = 1
Covered later in calculus; it follows from the line above by substitution
Summary
Definition of a limit
limx→a f(x) = L ⟺ left-hand limit = right-hand limit = L
A limit exists only when both sides approach the same value
🎯 Exam Points
①If the left- and right-hand limits differ, there is no limit
②f(a) and the limit are different ideas; the limit can exist even without f(a)
③For a 0/0 form, cancel by factoring or rationalizing, then substitute
④The rules for sums, products and quotients apply only when each limit exists
⑤If the denominator's limit is 0 in a quotient, it needs separate treatment
Worked Examples & Past Exam
Example 1
Evaluate limx→2 (x² − 4)/(x − 2).
1
Putting in x = 2 gives 0/0, so factor the numerator and cancel.
x2 - 4x - 2 = (x+2)(x-2)x-2 = x + 2
2
Put x = 2 into the simplified x + 2.
limx→2 (x + 2) = 4
▸ 4
The cancelling is valid only for x ≠ 2, but a limit looks only at values near x = 2, never at 2 itself, so there is no problem.
Example 2
Evaluate limx→0 (sin 3x)/x.
1
To use sin x / x → 1, the denominator must match the 3x inside the sine.
sin 3xx = 3 × sin 3x3x
2
As x → 0, 3x → 0, so sin 3x / 3x → 1.
3 × 1 = 3
▸ 3
The limit is 1 only when the expression inside the sine matches the denominator, which is why the fraction is multiplied and divided by 3.
2023 CSAT Math type, adapted
What is limx→1 (x² + 2x − 3)/(x − 1)?
①4
②0
③2
④3
⑤no limit
▸ ① 4
1
Putting in x = 1 gives 0/0, so factor the numerator as (x + 3)(x − 1).