Limit of a Function and One-Sided Limits

The limit of a function f at a point a is the value L that f(x) approaches as x gets close to a (but x ≠ a); one-sided limits describe the approach from the left (x → a⁻) or from the right (x → a⁺), and the limit exists exactly when both agree. In brief: lim_x→2 (x² − 4)/(x − 2) = 4, although the function is not defined at 2. For f(x) = |x|/x the left limit at 0 is −1 and the right limit is 1, so the limit does not exist.

The limit tells what the function does near a point, not at the point itself. It is the basis of continuity and of the derivative. This guide gives the intuitive idea and the ε–δ definition, explains one-sided limits, shows how to compute limits by substitution, factoring and conjugates, covers piecewise functions and the cases where limits do not exist, with a numerical table.

What does a limit mean?

We write lim_x→a f(x) = L if f(x) can be made as close to L as we like by taking x sufficiently close to a, with x ≠ a. The value f(a) is irrelevant: the function may be undefined at a or have a different value there. For this to make sense, a must be an accumulation point of the domain.

Example: f(x) = (x² − 4)/(x − 2) is not defined at x = 2, but for x ≠ 2 it equals x + 2. As x approaches 2, f(x) approaches 4, so the limit is 4.

What is the ε–δ definition?

lim_x→a f(x) = L means: for every ε > 0 there exists δ > 0 such that 0 < |x − a| < δ implies |f(x) − L| < ε. The number ε is the allowed error in the output and δ is how close the input must be.

Example 1: a linear function

Show that lim_x→3 (2x + 1) = 7. |f(x) − 7| = |2x − 6| = 2|x − 3|. For ε > 0 choose δ = ε/2. If |x − 3| < δ, then |f(x) − 7| < 2δ = ε ✓.

Example 2: a quadratic function

Show that lim_x→2 x² = 4. |x² − 4| = |x − 2||x + 2|. If |x − 2| < 1, then 1 < x < 3 and |x + 2| < 5. So |x² − 4| < 5|x − 2|. Choose δ = min(1, ε/5). Then |x − 2| < δ gives |x² − 4| < ε ✓. For ε = 0.05: δ = 0.01.

What are one-sided limits?

The limit from the right, lim_x→a⁺ f(x), uses only x > a; the limit from the left, lim_x→a⁻ f(x), uses only x < a. Then

lim_x→a f(x) = L if and only if lim_x→a⁻ f(x) = L and lim_x→a⁺ f(x) = L.

Example 3: piecewise function

f(x) = x + 1 for x < 2 and f(x) = 5 − x for x ≥ 2. Left limit at 2: 2 + 1 = 3. Right limit: 5 − 2 = 3. The limit exists and equals 3. Also f(2) = 3, so f is continuous at 2.

Example 4: a jump

g(x) = |x|/x. For x > 0, g(x) = 1; for x < 0, g(x) = −1. The right limit at 0 is 1 and the left limit is −1. They differ, so lim_x→0 g(x) does not exist.

Example 5: the floor function

⌊x⌋ at x = 3: from the left, the values are 2, so the left limit is 2; from the right, the values are 3, so the right limit is 3. At non-integer points the limit exists.

Example 6: an infinite one-sided limit

1/(x − 1): as x → 1⁺, the values grow without bound, so the right limit is +∞; as x → 1⁻, the left limit is −∞. The limit does not exist (as a real number).

How do you compute limits?

1. Direct substitution: if f is a polynomial, rational function (with nonzero denominator) or other elementary function defined at a, the limit equals f(a). For example lim_x→3 (x² + 2x) = 15.
2. Factoring and cancelling: for 0/0, factor and cancel the common factor.
3. Conjugates: for roots, multiply by the conjugate.

Example 7: factoring

lim_x→3 (x² − 9)/(x − 3) = lim (x + 3) = 6. And lim_x→1 (x² − 3x + 2)/(x² − 1) = lim (x − 2)(x − 1)/((x − 1)(x + 1)) = lim (x − 2)/(x + 1) = −1/2.

Example 8: a conjugate

lim_x→0 (√(x + 4) − 2)/x = lim [(x + 4) − 4]/(x(√(x + 4) + 2)) = lim 1/(√(x + 4) + 2) = 1/4.

Example 9: a complex fraction

lim_x→2 (1/x − 1/2)/(x − 2) = lim [(2 − x)/(2x)]/(x − 2) = lim −1/(2x) = −1/4. This limit is the derivative of 1/x at 2.

Example 10: a piecewise-defined parameter

For which k does lim_x→1 f(x) exist, with f(x) = kx + 2 for x < 1 and f(x) = x² + 3 for x ≥ 1? The right limit is 4. The left limit is k + 2. They agree when k = 2.

When does a limit not exist?

• Jump: the one-sided limits differ (|x|/x at 0).
• Infinite behaviour: the values grow without bound (1/x² → ∞ is an infinite limit; 1/x has no limit at 0).
• Oscillation: sin(1/x) as x → 0 oscillates between −1 and 1 infinitely often. Taking x = 1/(nπ) gives sin = 0, and x = 1/(π/2 + 2nπ) gives sin = 1, so two sequences approach 0 with different limits of f.

Numerical table for (x² − 4)/(x − 2)

From both sides the values approach 4, so the limit is 4. A table suggests the limit, but a proof needs algebra or the definition.

Where does the definition come from?

Newton and Leibniz used the idea of limits informally in the seventeenth century, Newton speaking of “ultimate ratios”. The modern definition with ε and δ goes back to Cauchy (1821), who defined limits with inequalities, and was made fully rigorous by Karl Weierstrass in the 1850s–1860s. Before this, calculus rested on infinitely small quantities that mathematicians such as Berkeley criticised in the eighteenth century.

Where are limits used?

Limits define the instantaneous speed as the limit of average speeds, the tangent slope as the limit of secant slopes and the area under a curve as a limit of sums. In engineering, the behaviour of a system near a critical value is analysed by limits; in economics, marginal cost is the limit of the cost change per unit.

Which mistakes are common?

Confusing the limit with the function value

The limit at a does not depend on f(a). For f(x) = x + 2 for x ≠ 2 and f(2) = 10, the limit at 2 is still 4.

Checking only one side

Both one-sided limits must exist and be equal.

Cancelling without factoring

Only factors can be cancelled: (x + 2)/(x + 5) cannot be simplified.

Believing that a table is a proof

Numerical evidence can mislead: sin(π/x) takes the value 0 at all x = 1/n, but its limit at 0 does not exist.