The Epsilon-Delta Definition of a Limit
State what a limit claims with no appeal to motion: for every tolerance around the limiting value there is a distance around the point that keeps outputs inside it. Prove a simple limit by producing that distance.
What a learner can do afterwards
- Interpret the tolerance, input bound and quantifier order in an epsilon-delta statement
- Calculate and check input-output error bounds for linear and quadratic functions, including counterexamples to proposed bounds
- Explain why a limit proof must cover arbitrarily small positive tolerances
1 · Read
A finite limit claim fixes an input point a and target L. The domain must contain inputs other than a arbitrarily close to a. Epsilon is a positive tolerance for the output error |f(x)-L|. A challenger may choose any epsilon. You then choose a positive input distance delta that keeps every permitted output within epsilon of L.
Assume domain inputs other than a occur arbitrarily close to a. The limit is L if, for every epsilon > 0, some delta > 0 makes every domain input x with 0 < |x-a| < delta satisfy |f(x)-L| < epsilon. Delta may depend on epsilon, a and the function, but not on the individual input x. One delta must work for all permitted inputs.
Choose delta after epsilon, then test all inputs covered by that choice.
For f(x) = mx+b with m > 0 and L = ma+b, the output error is m|x-a|. An input distance below delta gives output error below m delta. Thus delta = epsilon/m works, and every smaller positive delta works too. Given delta, m delta is the boundary output-error bound.
The condition 0 < |x-a| < delta means a-delta < x < a+delta, with x = a excluded. The endpoints are excluded too because the inequalities are strict. The value f(a) can be missing or different from L without changing a limit. For example, 0 < |x-6| < 0.4 excludes 6, 5.6 and 6.4.
For a line with nonzero slope m, use its absolute value: |f(x)-L| = |m||x-a|, so delta = epsilon/|m| works. Smaller positive distances work too. A negative slope still gives a positive error. If m = 0, f is constant and L = b. The output error is zero, so every positive delta works; do not divide by zero.
The absolute slope scales the distance from the input point.
Checking one tolerance does not prove a limit. A smaller epsilon may need a smaller delta. However, a delta that works for one epsilon also works for every larger epsilon, since the output requirement becomes looser. To prove a limit, give a working delta for an arbitrary positive epsilon.
For f(x) = x² near 0 with L = 0, the output error is x². The inequality x² < epsilon is equivalent to |x| < sqrt(epsilon), so delta = sqrt(epsilon) works. It is the largest working delta. Any larger delta includes x = sqrt(epsilon), whose output error equals epsilon and fails the strict requirement.
Prove that 5x-2 tends to 3 at x = 1. Let epsilon > 0 and choose delta = epsilon/5. Assume 0 < |x-1| < delta. Then |(5x-2)-3| = 5|x-1| < 5 delta = epsilon. This proves the claim for every positive epsilon. The proof uses a strict inequality, not just equality at the boundary.
Epsilon is the demanded closeness, delta is your answer, and the pair must work for every epsilon.
Original epsilon-delta lesson, questions and flashcards written for this learning dataset.
The epsilon-delta practice questions, reading and flashcards were written for this learning dataset.
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