Building intuitive understanding of limits as the foundation of calculus. Covers the informal definition of a limit (the value a function approaches as x approaches a given value), evaluating limits from tables and graphs, one-sided limits, limits at infinity and horizontal asymptotes, and cases where limits do not exist.
High School Advanced • 9-12
The concept of a limit is the gateway to calculus and one of the most profound ideas in all of mathematics. Limits formalize the intuition of "approaching" a value. They answer the question: What value does a function get closer and closer to as the input gets closer and closer to some target? Remarkably, the function does not need to actually reach or even be defined at that target -- what matters is the trend.
We write:
and read this as "the limit of f(x) as x approaches c equals L." This means: as x gets arbitrarily close to c (from both sides), f(x) gets arbitrarily close to L.
Critical insight: the limit depends on what happens near c, not at c. The function f(c) might not exist, or it might exist but differ from L. The limit only cares about the trend as x approaches c.
To estimate a limit numerically, compute f(x) for values of x approaching c from both sides:
| x | 1.9 | 1.99 | 1.999 | → 2 ← | 2.001 | 2.01 | 2.1 |
|---|---|---|---|---|---|---|---|
| f(x) | 3.61 | 3.9601 | 3.996 | ? | 4.004 | 4.0401 | 4.41 |
If f(x) = x^2, the table shows f(x) approaching 4 as x approaches 2 from both sides. We conclude lim (x → 2) x^2 = 4.
On a graph, the limit as x → c is the y-value that the curve approaches as you trace along it toward x = c. Look at the curve's behavior near c, not the dot (or hole) at c itself. A filled dot at a different height than the curve indicates f(c) differs from the limit.
Sometimes the function approaches different values from the left and right:
The two-sided limit exists if and only if both one-sided limits exist and are equal:
We can also ask what happens as x grows without bound:
This means f(x) approaches L as x becomes very large. If such a limit L exists (and is finite), the line y = L is a horizontal asymptote.
For rational functions (ratios of polynomials), limits at infinity depend on the degrees of the numerator and denominator:
A limit fails to exist in three main situations:
Problem: Find lim (x → 3) (x^2 - 9)/(x - 3).
Solution: Direct substitution gives 0/0, which is indeterminate. Factor the numerator:
Now the limit is straightforward:
The function is not defined at x = 3 (there is a hole), but the limit exists and equals 6.
Problem: For the function f(x) = { x + 1 if x < 2; 5 if x = 2; 2x - 1 if x > 2 }, find the one-sided limits and determine if the two-sided limit exists at x = 2.
Solution:
Left-hand limit: lim (x → 2−) (x + 1) = 3
Right-hand limit: lim (x → 2+) (2x - 1) = 3
Both one-sided limits equal 3, so lim (x → 2) f(x) = 3.
Note: f(2) = 5 ≠ 3. The limit exists but does not equal the function value. The function is discontinuous at x = 2.
Problem: Find lim (x → ∞) (3x^2 + 5x)/(2x^2 - 1).
Solution: The degrees of numerator and denominator are both 2 (equal). The limit is the ratio of leading coefficients:
To see why, divide every term by x^2 (the highest power):
The horizontal asymptote is y = 3/2.
Limits are not just a theoretical curiosity. The entire edifice of calculus rests on two limits: the derivative (a limit of difference quotients, measuring instantaneous rate of change) and the integral (a limit of Riemann sums, measuring accumulated area). Understanding limits now will prepare you for the most powerful mathematical tools ever developed.
Problem 1: Find lim (x → 5) (x^2 - 25)/(x - 5).
Factor: (x^2 - 25)/(x - 5) = (x - 5)(x + 5)/(x - 5) = x + 5 for x ≠ 5.
lim (x → 5) (x + 5) = 10.
Problem 2: Find lim (x → ∞) (5x + 3)/(2x^2 + 1).
Degree of numerator (1) < degree of denominator (2), so the limit is 0.
Alternatively, divide by x^2: (5/x + 3/x^2)/(2 + 1/x^2) → (0 + 0)/(2 + 0) = 0.
Problem 3: For f(x) = { 2x if x < 1; 4 - x if x ≥ 1 }, find lim (x → 1) f(x) and determine if f is continuous at x = 1.
Left-hand limit: lim (x → 1−) 2x = 2.
Right-hand limit: lim (x → 1+) (4 - x) = 3.
Since 2 ≠ 3, the two-sided limit does not exist. Therefore f is not continuous at x = 1.
Problem 4: Does lim (x → 0) 1/x^2 exist? Explain.
As x → 0 from either side, 1/x^2 grows without bound (goes to +∞).
Since the function does not approach a finite value, the limit does not exist. We can write lim (x → 0) 1/x^2 = +∞ (informally), but this means the limit fails to exist in the finite sense. There is a vertical asymptote at x = 0.
Problem 5: Find lim (x → -2) (x^2 + 5x + 6)/(x + 2).
Factor: x^2 + 5x + 6 = (x + 2)(x + 3).
(x + 2)(x + 3)/(x + 2) = x + 3 for x ≠ -2.
lim (x → -2) (x + 3) = 1.
A limit describes the value a function approaches as the input approaches a target, regardless of the function's actual value (or existence) at that target. One-sided limits examine the approach from a single direction, and the two-sided limit exists only when both one-sided limits agree. Limits at infinity reveal horizontal asymptotes. Limits fail to exist when one-sided limits disagree, when the function grows unboundedly, or when it oscillates without settling. Mastering limits opens the door to calculus and the precise study of change and accumulation.