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Daily Math Minute

Unit 1: Limits & Continuity

Evaluating Limits

Estimating and evaluating limits using graphical, numerical, and algebraic methods.

Advanced25 min lesson3 min readUpdated August 12, 2026Author not yet attributed

Reading a Function's Approach, Not Its Value

f(x) = (x² − 4)/(x − 2) is undefined at x = 2 — direct substitution gives 0/0. Before reading on: does that undefined value mean nothing can be said about how f behaves right around x = 2?

Definition — Limit

limₓ→c f(x) is the value f(x) approaches as x gets arbitrarily close to c from both sides, independent of whether f(c) itself is defined. A limit fails to exist when the left-hand and right-hand approaches disagree, or when f grows without bound near c.

Worked Example — Estimating and Confirming a Limit

Estimate limₓ→2 (x² − 4)/(x − 2). Numerically: f(1.9) = 3.9, f(1.99) = 3.99, f(2.01) = 4.01, f(2.1) = 4.1 — approaching 4 from both sides. Algebraically, factoring confirms it: (x − 2)(x + 2)/(x − 2) = x + 2 for x ≠ 2, and x + 2 at x = 2 is 4. Both methods agree: the limit is 4, even though f(2) itself is undefined.

Function Explorer

Transform: g(x) = a·f(b(x − h)) + k
1
1
0
0
Composition
Analysis (of the transformed function, in view)
y-intercept
(0, 0)
x-intercepts
(-9.42, 0), (-6.28, 0), (-3.14, 0), (0, 0), (3.14, 0), (6.28, 0), (9.42, 0)
Extrema
local min at (-7.85, -1); local max at (-4.71, 1); local min at (-1.57, -1); local max at (1.57, 1); local min at (4.71, -1); local max at (7.85, 1)
Inflection points
(-9.42, 0), (-6.28, 0), (-3.14, 0), (0, 0), (3.14, 0), (6.28, 0), (9.42, 0)
Vertical asymptotes
none found in view
Horizontal asymptotes
none found
Domain
all real numbers in view
Range (estimated)
approximately [-1, 1]

Worked Example — Resolving 0/0 by Rationalizing

Evaluate limₓ→0 (√(x + 9) − 3)/x. Direct substitution gives 0/0. Multiply by the conjugate: [(x + 9) − 9] / [x(√(x + 9) + 3)] = x / [x(√(x + 9) + 3)] = 1/(√(x + 9) + 3), valid for x ≠ 0. As x → 0: 1/(3 + 3) = 1/6. Check numerically: at x = 0.001, (√9.001 − 3)/0.001 ≈ 0.1667, matching 1/6.

Worked Example — Applying Limit Laws to Given Values

Given limₓ→1 f(x) = 4 and limₓ→1 g(x) = 2, find limₓ→1 [f(x)g(x) − 3f(x)]. By the product, difference, and constant-multiple laws: (4)(2) − 3(4) = 8 − 12 = −4 — no formula for f or g is needed.

Tip

Try direct substitution first. Reach for factoring or rationalizing only when it produces an indeterminate form like 0/0 — most limits need no algebraic rewriting at all.

Common Mistakes

  • Concluding a limit doesn't exist just because the function is undefined at that exact point.

    Being undefined at c says nothing by itself about limₓ→c f(x) — (x² − 4)/(x − 2) is undefined at x = 2 but still has a limit of 4 there.

  • Substituting into the original, unsimplified expression after factoring or rationalizing.

    Once an indeterminate form is resolved algebraically, substitute into the simplified expression — that's the entire point of the algebra.

Key Takeaways

  • A limit describes what a function approaches, independent of its actual value at that point.
  • 0/0 on direct substitution doesn't mean a limit fails to exist — factoring or rationalizing often resolves it.
  • Limit laws let a limit be evaluated from given values alone, without knowing the underlying functions.

Summary

Limits describe local approach; the next lesson uses that idea to define exactly what makes a function continuous.

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