Epsilon-delta limits
Objective. Construct rigorous function-limit proofs.
- 1. Learn
- 2. Worked example
- 3. Practice
- 4. Feedback
- 5. Continue
Learn: the key idea
Epsilon-delta limits give a rigorous definition of lim(x→a) f(x) = L: for every ε > 0 there exists δ > 0 such that 0 < |x − a| < δ implies |f(x) − L| < ε. The method is to bound |f(x) − L| in terms of |x − a| using algebra, then choose δ small enough to force the bound below ε. For f(x) = 2x + 1 at a = 3, |f(x) − 7| = 2|x − 3|, so choosing δ = ε/2 guarantees |f(x) − 7| < ε whenever |x − 3| < δ. This formalises the intuitive idea that f(x) gets arbitrarily close to L as x approaches a.
Worked example
Using an epsilon-delta argument, prove the limit of f(x) = 3x − 2 as x → 4 equals 10.
- 1Compute |f(x) − 10| = |3x − 2 − 10| = |3x − 12| = 3|x − 4|.
- 2We require 3|x − 4| < ε, which holds when |x − 4| < ε/3.
- 3Choose δ = ε/3, so 0 < |x − 4| < δ forces |f(x) − 10| < ε.
- 4Since ε was arbitrary, the limit definition is satisfied for every ε > 0.
Answer: δ = ε/3 proves lim(x→4)(3x − 2) = 10
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Practice
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Start practisingCommon mistake. A common error is choosing δ before bounding the expression, so the inequality |f(x) − L| < ε is never actually guaranteed. Always factor out |x − a| first, then pick δ as a function of ε.
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