Introductory Limits

A limit describes the value a function approaches as its input approaches a target. It is the bridge from pre-calculus to calculus. Review domains and graph behaviour in Functions and removable holes in Polynomial and Rational Functions first.

Approaching Is Not Arriving

The notation limx→a f(x) = L says that f(x) can be made as close as desired to L by taking x sufficiently close to a. The function value f(a) may equal L, differ from it, or be undefined.

A graph approaching the same height from both sides of a hole A straight line rises from left to right and has an open circle at x equals two, y equals four. Arrows on both sides point toward the open circle. A filled point at x equals two, y equals one shows that the function value differs from the limit. hole (2, 4) f(2) = 1 x y
Both sides approach 4, so the limit is 4 even though the defined point gives f(2) = 1.

Estimating from a Table

For f(x) = (x2 − 4)/(x − 2), the expression is undefined at 2. Nearby values still reveal its limit.

x1.91.992.012.1
f(x)3.93.994.014.1

Values from both sides approach 4. A table supports an estimate; algebra explains it exactly.

Limit Laws and Direct Substitution

Limits respect sums, differences, products, powers, and quotients when the denominator limit is not zero. For polynomials, substitute directly.

Worked example: direct substitution

lim as x→3 of (2x² − x + 4)
= 2(3)² − 3 + 4
= 18 − 3 + 4
= 19

Indeterminate Forms and Factoring

Substitution that produces 0/0 does not mean the limit is zero or nonexistent. It signals that the expression may simplify for nearby inputs.

Worked example: remove a common factor

lim as x→2 of (x² − 4)/(x − 2)

= lim as x→2 of [(x − 2)(x + 2)]/(x − 2)
= lim as x→2 of (x + 2), for x ≠ 2
= 2 + 2
= 4

Cancelling is legitimate in a limit because we examine nearby values with x ≠ 2.

One-Sided Limits and Existence

The left-hand limit uses values less than a, written x → a−. The right-hand limit uses values greater than a, written x → a+. A two-sided limit exists only when both one-sided limits exist and are equal.

Worked example: a jump

Let f(x) = 1 for x < 0 and f(x) = 3 for x ≥ 0.

lim as x→0⁻ of f(x) = 1
lim as x→0⁺ of f(x) = 3

Because 1 ≠ 3, lim as x→0 of f(x) does not exist.
The defined value f(0) = 3 does not change that conclusion.

Infinite Limits

If values grow without bound near a vertical asymptote, write an infinite limit. Infinity describes behaviour; it is not a real number.

For f(x) = 1/(x − 2):
lim as x→2⁻ of f(x) = −∞
lim as x→2⁺ of f(x) = +∞
Therefore the two-sided limit does not exist.

Practice Set

  1. Evaluate limx→4 (3x − 7).
  2. Evaluate limx→5 (x2 − 25)/(x − 5).
  3. If the left-hand limit at x = 2 is 6 and the right-hand limit is 6, but f(2) = 9, state the two-sided limit.
  4. If the left-hand limit is −1 and the right-hand limit is 4, does the two-sided limit exist?
  5. State both one-sided limits of 1/x as x approaches 0.

Answer Checks

  1. 3(4) − 7 = 5.
  2. Factor to (x − 5)(x + 5)/(x − 5), then substitute: 5 + 5 = 10.
  3. The two-sided limit is 6. The value at the point can differ from the limit.
  4. No. Unequal one-sided limits mean the two-sided limit does not exist.
  5. limx→0− 1/x = −∞ and limx→0+ 1/x = +∞.