Calculus I · Limits and continuity

Prerequisites: functions, graphs, interval notation and factoring. By the end you can read one-sided limits, evaluate finite limits, test continuity, justify a limit with an ε–δ proof, and apply the Intermediate Value Theorem.

1. Nearby values and one-sided limits

A limit describes where the outputs go as inputs approach a point through the domain. The value at that point may be different or undefined. The left-hand limit uses inputs below the point, and the right-hand limit uses inputs above it. When inputs are available arbitrarily close on both sides, a finite two-sided limit exists exactly when the two one-sided limits exist and agree.

2. Compute limits and prove them

For finite limits, sums and products combine in the expected way. A quotient law also works when the denominator limit is nonzero. Direct substitution works for polynomials and for rational functions where their denominator is nonzero. If substitution yields 0/0, simplify for nearby inputs first, then evaluate the new expression.

Worked example · A removable hole

f(x) = (x² − 4)/(x − 2) = x + 2 (x ≠ 2)

Factor the numerator and cancel x − 2 only when x ≠ 2. Then the limit at 2 is 4. Defining f(2) = 4 fills the hole and makes the extended function continuous at 2.

Proof · The ε–δ definition

For every output tolerance ε > 0, choose an input distance δ > 0 such that every domain point with 0 < |x − a| < δ satisfies |f(x) − L| < ε. To prove the limit of 3x + 1 at x = 2 is 7, use δ = ε/3. Then |(3x + 1) − 7| = 3|x − 2| < 3δ = ε.

3. Continuity and the Intermediate Value Theorem

Continuity at an interior point requires a defined function value, an existing finite limit, and equality of the two. On a closed interval use the appropriate one-sided limit at each endpoint. If a function is continuous on [a,b], it takes every value between f(a) and f(b) somewhere in that interval. A sign change guarantees a root inside, while uniqueness needs an additional argument such as strict monotonicity.

Engineering example · Switching a sensor range

A sensor model uses V(t) = 2t for t < 1 and V(t) = kt + 1 for t ≥ 1. The left limit at 1 is 2 V. The right limit and actual value are k + 1 V. Set k = 1 for continuity. With k = 3 there is a jump from 2 V to 4 V, so a threshold of 3 V need not be reached during the switch.

Common mistakes

• Treating 0/0 as an answer. It signals that substitution alone is inconclusive.
• Using the value at a point to infer its limit, or checking only one side of a jump.
• Using the Intermediate Value Theorem across a discontinuity, or using it to claim uniqueness.

Practice with worked solutions

1. Find the limit of (x² − 9)/(x − 3) at 3 and give a continuous extension.
2. For x ≠ 0, find the one-sided limits of |x|/x at zero.
3. Prove with ε and δ that 5x − 2 tends to 3 as x tends to 1.

Mini-exam · 20 minutes · 10 points

1. Evaluate the limit of (x² + x − 6)/(x − 2) at 2. Show your algebra. (3 points)
2. For f(x) = x + 1 when x < 2 and f(x) = cx − 1 when x ≥ 2, find c that makes f continuous at 2. (4 points)
3. Use the Intermediate Value Theorem to prove that x³ + x − 3 has a zero in (1,2). (3 points)

Free readings and next lesson

• OpenStax §2.2 · Limits of a function
• OpenStax §2.3 · Limit laws
• OpenStax §2.4 · Continuity
• OpenStax §2.5 · Precise definition

Next in Calculus I: Rates of Change