A limit asks what value a function gets close to as x gets close to some number. It does not ask what happens exactly at that number, which is the whole point: limits let you talk about places where a function is undefined, and they are the foundation of both derivatives and integrals.
What a limit means
The notation
limx→a f(x) = L
reads “the limit of f(x) as x approaches a is L”. It means that as x gets closer and closer to a, from both sides, f(x) gets closer and closer to L.
Take f(x) = (x² − 1)/(x − 1). At x = 1 it is 0/0, which is undefined. But look at nearby values:
| x | 0.9 | 0.99 | 0.999 | 1 | 1.001 | 1.01 | 1.1 |
|---|---|---|---|---|---|---|---|
| f(x) | 1.9 | 1.99 | 1.999 | undefined | 2.001 | 2.01 | 2.1 |
The outputs close in on 2 from both sides, so limx→1 f(x) = 2, even though f(1) doesn’t exist. The graph is the line y = x + 1 with a hole at (1, 2).
Method 1: direct substitution
If the function is a polynomial, or any function that is continuous at a, the limit is just f(a). Always try this first.
Example Find limx→3 (x² + 2x − 4)
Substitute: 9 + 6 − 4 = 11.
If substitution gives a real number, you are done. If it gives a nonzero number over zero, like 5/0, the limit does not exist as a finite number (the graph has a vertical asymptote there). If it gives 0/0, that is called an indeterminate form, and you need one of the methods below.
Method 2: factor and cancel
0/0 from a fraction of polynomials means both the top and bottom have a factor of (x − a). Factor, cancel it, then substitute.
Example Find limx→4 (x² − 16)/(x − 4)
- Substitution gives 0/0.
- Factor the top: (x − 4)(x + 4)/(x − 4).
- Cancel: x + 4.
- Substitute: 4 + 4 = 8.
Example Find limx→−2 (x² + 5x + 6)/(x² − 4)
- Substitution gives 0/0.
- Factor: (x + 2)(x + 3) / ((x + 2)(x − 2)).
- Cancel: (x + 3)/(x − 2).
- Substitute: 1/(−4) = −1/4.
The factoring guide covers every factoring pattern you will need here.
Method 3: rationalize
When a square root causes the 0/0, multiply the top and bottom by the conjugate.
Example Find limx→9 (√x − 3)/(x − 9)
- Substitution gives 0/0.
- Multiply top and bottom by √x + 3: (x − 9) / ((x − 9)(√x + 3)).
- Cancel: 1/(√x + 3).
- Substitute: 1/(3 + 3) = 1/6.
One-sided limits
limx→a⁻ means x approaches a from the left (values below a). limx→a⁺ means from the right. The two-sided limit exists only if both one-sided limits exist and are equal.
Example f(x) = x + 1 for x < 2, and f(x) = 5 − x for x ≥ 2. Does limx→2 f(x) exist?
- From the left: 2 + 1 = 3.
- From the right: 5 − 2 = 3.
- They match, so the limit is 3.
Example Find limx→0 |x|/x
From the right, |x|/x = x/x = 1. From the left, |x|/x = −x/x = −1. The one-sided limits differ, so the limit does not exist.
Limits at infinity
limx→∞ asks what happens as x grows without bound. For a fraction of polynomials, compare the highest powers on top and bottom:
| Situation | Limit as x → ∞ | Example |
|---|---|---|
| Bottom has the higher power | 0 | (3x + 1)/(x² − 5) → 0 |
| Same highest power | ratio of leading coefficients | (4x² − x)/(2x² + 7) → 2 |
| Top has the higher power | grows without bound (∞ or −∞) | (x³ + 1)/(x² + 1) → ∞ |
A finite limit at infinity is a horizontal asymptote of the graph.
Example Find limx→∞ (6x³ − 2x)/(3x³ + x² + 1)
Both have x³ as the highest power. The limit is 6/3 = 2, so y = 2 is a horizontal asymptote.
Two special limits
These come up constantly in calculus, especially when deriving the derivatives of sin x and eˣ:
- limx→0 (sin x)/x = 1 (x in radians)
- limx→0 (1 − cos x)/x = 0
Continuity
A function is continuous at x = a if three things are true:
- f(a) is defined.
- limx→a f(x) exists.
- They are equal.
Informally, you can draw the graph through that point without lifting your pencil. The function (x² − 1)/(x − 1) from the start fails test 1 at x = 1: it has a hole. Defining f(1) = 2 would “fill the hole” and make it continuous.
How limits lead to derivatives
The derivative is defined as a limit:
f′(x) = limh→0 [f(x + h) − f(x)]/h
Substituting h = 0 always gives 0/0, so finding a derivative from this definition is exactly the factor-and-cancel method above. For f(x) = x², the fraction simplifies to 2x + h, and the limit as h → 0 is 2x.
Try a few
Find limx→2 (3x² − x + 1).
Show answer
11. Substitute: 12 − 2 + 1.
Find limx→5 (x² − 25)/(x − 5).
Show answer
10. Factor to (x + 5)(x − 5)/(x − 5) = x + 5, then substitute.
Find limx→1 (x² + 2x − 3)/(x² − 1).
Show answer
2. Factor: (x + 3)(x − 1)/((x + 1)(x − 1)) = (x + 3)/(x + 1), which is 4/2 at x = 1.
Find limx→0 (√(x + 4) − 2)/x.
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1/4. Multiply by the conjugate to get 1/(√(x + 4) + 2), which is 1/4 at x = 0.
Find limx→∞ (5x² + 3)/(2x² − x).
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5/2. Same highest power, so take the ratio of leading coefficients.
Find limx→3 1/(x − 3).
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It does not exist. From the right the values go to +∞, and from the left to −∞.