DWG NO. 10.1 — Lesson 1 of 5

The Intuitive Idea of a Limit

Unit 10: Limits and an Introduction to Calculus · ~20 min

Every function you've studied so far, you evaluated by plugging in a number. But some of the most useful questions in mathematics are about what a function is headed toward as the input gets close to a value — not what happens exactly there. That question is a limit, and it's the foundation the rest of this unit, and all of calculus, is built on.

Notation and informal meaning

The expression li\(m_{x→a}\) f(x) = L is read "the limit of f(x) as x approaches a equals L." It means: as x gets arbitrarily close to a — from either side, but never actually equal to a — the output f(x) gets arbitrarily close to the single number L. Crucially, this says nothing about f(a) itself. f might be undefined right at x = a, or defined with some completely different value there, and the limit wouldn't care.

Estimating a limit from a table

One reliable way to estimate a limit is to plug in x-values that creep closer and closer to a from both sides and watch what f(x) approaches. If the values from the left and the values from the right are both closing in on the same number, that number is the limit.

One-sided limits

Sometimes it's useful to only approach from one direction. li\(m_{x→a^{-}}\) f(x) is the left-hand limit (x approaches a through values less than a); li\(m_{x→a^{+}}\) f(x) is the right-hand limit (x approaches through values greater than a). The two-sided limit li\(m_{x→a}\) f(x) exists if and only if both one-sided limits exist and are equal to each other.

• Limit exists: left-hand limit = right-hand limit = some number L. Then li\(m_{x→a}\) f(x) = L.
• Limit does not exist (DNE): the left-hand and right-hand limits disagree, or either one fails to settle on a single number (e.g. the function oscillates or blows up).

Guided practice

• 1.Use a table of values (x = 1.9, 1.99, 2.01, 2.1) to estimate li\(m_{x→2}\) (x²−4)/(x−2).
  Show answer

  f(1.9)=3.9, f(1.99)=3.99, f(2.01)=4.01, f(2.1)=4.1 — both sides approach 4. lim = 4 (this simplifies to x+2 everywhere except x=2).
• 2.Let f(x) = 2x for x < 0, and f(x) = x + 3 for x ≥ 0. Find the left-hand and right-hand limits at x = 0, and state whether li\(m_{x→0}\) f(x) exists.
  Show answer

  Left: li\(m_{x→0^{-}}\) 2x = 0. Right: li\(m_{x→0^{+}}\) (x+3) = 3. They disagree, so li\(m_{x→0}\) f(x) does not exist.
• 3.Does a limit li\(m_{x→a}\) f(x) require f(a) to be defined? Explain briefly.
  Show answer

  No. A limit only describes what f(x) approaches as x gets close to a — it never actually evaluates f at a, so f(a) can be undefined and the limit can still exist.
• 4.Estimate li\(m_{x→2}\) (x²+x−6)/(x−2) using a table of values near x = 2.
  Show answer

  f(1.9)=4.9, f(1.99)=4.99, f(2.01)=5.01, f(2.1)=5.1 — both sides approach 5. lim = 5 (this simplifies to x+3 everywhere except x=2).
• 5.Let g(x) = 3 for every x ≠ 1, and g(1) = 0. Find li\(m_{x→1}\) g(x).
  Show answer

  The limit only cares about values near x = 1, not at x = 1 itself — and g(x) = 3 for every x near 1. li\(m_{x→1}\) g(x) = 3, even though g(1) = 0.