Mathematics.

limits

Epsilon-Delta Definition of Limits

Calculus I50 minDifficulty5 out of 10

You should know: limit

Overview

The epsilon-delta definition makes the intuitive notion of a limit precise. We say lim_{x->a} f(x) = L if for every epsilon > 0 (no matter how small), there exists delta > 0 such that if 0 < |x - a| < delta then |f(x) - L| < epsilon. This formalizes 'f(x) gets arbitrarily close to L as x approaches a'. The epsilon-delta approach, developed by Cauchy and Weierstrass in the 19th century, replaced vague geometric intuition with rigorous inequalities and is the foundation of real analysis.

Intuition

Challenge-response game: your opponent picks epsilon > 0 (a small tolerance around L). You win if you can find delta > 0 such that whenever x is within delta of a (but not equal to a), f(x) is within epsilon of L. You must win for every possible epsilon your opponent chooses. If you can always win, the limit exists and equals L. The key: 0 < |x-a| < delta (x is close to a but not equal) implies |f(x)-L| < epsilon (f(x) is close to L).

Formal Definition

Definition

lim_{x->a} f(x) = L means: for every real epsilon > 0, there exists delta > 0 such that for all x in the domain of f: 0 < |x - a| < delta implies |f(x) - L| < epsilon. The condition 0 < |x-a| says x != a (we don't require f(a) = L or even f(a) to be defined). To prove a limit, one typically: (1) guess L; (2) start with |f(x)-L| < epsilon; (3) work backwards to find delta in terms of epsilon; (4) write the proof forward: assume 0 < |x-a| < delta, then derive |f(x)-L| < epsilon.

limxaf(x)=L    ε>0  δ>0:  0<xa<δf(x)L<ε\lim_{x \to a} f(x) = L \iff \forall \varepsilon > 0\; \exists \delta > 0:\; 0 < |x-a| < \delta \Rightarrow |f(x)-L| < \varepsilon
Epsilon-delta definition
f(x)L<ε (output tolerance),0<xa<δ (input restriction)|f(x)-L| < \varepsilon \text{ (output tolerance)},\quad 0 < |x-a| < \delta \text{ (input restriction)}
Geometric interpretation
limxaf(x)L    ε>0:  δ>0  x:0<xa<δ and f(x)Lε\lim_{x\to a} f(x) \ne L \iff \exists \varepsilon > 0:\; \forall \delta > 0\; \exists x: 0 < |x-a| < \delta \text{ and } |f(x)-L| \ge \varepsilon
Limit does NOT equal L

Notation

NotationMeaning
ε>0\varepsilon > 0Output tolerance (how close f(x) must be to L)
δ>0\delta > 0Input restriction (how close x must be to a)
xa|x-a|Distance from x to a
0<xa<δ0 < |x-a| < \deltax is near a but not equal to a

Theorems

Theorem 1: Uniqueness of Limits
Iflimx>af(x)=Landlimx>af(x)=M,thenL=M.Proof:supposeL!=M.Letepsilon=LM/2>0.Bythedefinitions,thereexistdelta1anddelta2suchthatif0<xa<delta1thenf(x)L<epsilon,andif0<xa<delta2thenf(x)M<epsilon.Letdelta=min(delta1,delta2).For0<xa<delta:LM<=Lf(x)+f(x)M<2epsilon=LM,acontradiction.If lim_{x->a} f(x) = L and lim_{x->a} f(x) = M, then L = M. Proof: suppose L != M. Let epsilon = |L-M|/2 > 0. By the definitions, there exist delta_1 and delta_2 such that if 0 < |x-a| < delta_1 then |f(x)-L| < epsilon, and if 0 < |x-a| < delta_2 then |f(x)-M| < epsilon. Let delta = min(delta_1, delta_2). For 0 < |x-a| < delta: |L-M| <= |L-f(x)| + |f(x)-M| < 2*epsilon = |L-M|, a contradiction.
Theorem 2: Algebra of Limits
Iflimx>af(x)=Landlimx>ag(x)=M,then:lim(f+g)=L+M;lim(fg)=LM;lim(f/g)=L/M(ifM!=0);lim(cf)=cL.Eachofthesefollowsfromtheepsilondeltadefinitionusingthetriangleinequalityandotheralgebraicmanipulations.If lim_{x->a} f(x) = L and lim_{x->a} g(x) = M, then: lim(f+g) = L+M; lim(f*g) = L*M; lim(f/g) = L/M (if M != 0); lim(c*f) = c*L. Each of these follows from the epsilon-delta definition using the triangle inequality and other algebraic manipulations.
Theorem 3: Epsilon-Delta Proof of a Polynomial Limit
Foranypolynomialp(x)andanyrealnumbera,limx>ap(x)=p(a).Proofsketch:bythealgebraoflimits,itsufficestoprovelimx>axn=an(basecasex1isdirect;inductionuseslim(fg)=lim(f)lim(g)).Forlimx>ax=a:givenepsilon>0,takedelta=epsilon.Thenxa<delta=epsilonimmediatelygivesxa<epsilon.For any polynomial p(x) and any real number a, lim_{x->a} p(x) = p(a). Proof sketch: by the algebra of limits, it suffices to prove lim_{x->a} x^n = a^n (base case x^1 is direct; induction uses lim(f*g) = lim(f)*lim(g)). For lim_{x->a} x = a: given epsilon > 0, take delta = epsilon. Then |x-a| < delta = epsilon immediately gives |x-a| < epsilon.

Worked Examples

  1. 1

    We need: for every epsilon > 0, find delta > 0 such that 0 < |x-2| < delta implies |(3x-1) - 5| < epsilon.

  2. 2

    Simplify: |(3x-1) - 5| = |3x - 6| = 3|x - 2|.

    3x6=3x2|3x - 6| = 3|x-2|
  3. 3

    We want 3|x-2| < epsilon, i.e., |x-2| < epsilon/3. So choose delta = epsilon/3.

  4. 4

    Proof: given epsilon > 0, let delta = epsilon/3. Assume 0 < |x-2| < delta = epsilon/3. Then |(3x-1)-5| = 3|x-2| < 3*(epsilon/3) = epsilon. Done.

    δ=ε3\delta = \frac{\varepsilon}{3}

✓ Answer

Choose delta = epsilon/3. Then 0 < |x-2| < delta implies |(3x-1)-5| = 3|x-2| < epsilon.

Practice Problems

Mediumproof writing

Use the epsilon-delta definition to prove lim_{x->4} sqrt(x) = 2.

Common Mistakes

Common Mistake

Confusing the limit with the value of the function: thinking lim_{x->a}f(x)=f(a) must be true.

The limit lim_{x->a}f(x) is about the behavior of f near a, NOT at a. The function need not be defined at a, or f(a) may differ from the limit. For example, f(x) = (x^2-1)/(x-1) has a limit of 2 as x->1, even though f(1) is undefined. The condition 0 < |x-a| in the definition explicitly excludes x=a.

Quiz

In the epsilon-delta definition lim_{x->a}f(x)=L, which is chosen FIRST?

Historical Background

Newton and Leibniz invented calculus in the 17th century using intuitive notions of infinitesimals. Euler's 18th century calculus was powerful but lacked rigorous foundations. Cauchy (1821) introduced the epsilon-delta language in his Cours d'analyse, though his formulation was still somewhat informal. Weierstrass (1860s) gave the modern precise definition. The need for rigor came from paradoxes: Dirichlet's 1829 function (1 on rationals, 0 on irrationals) required a precise definition to discuss its limits.

  1. 1821

    Cauchy introduces proto-epsilon-delta language in Cours d'analyse

    Augustin-Louis Cauchy

  2. 1860

    Weierstrass gives the modern precise epsilon-delta definition

    Karl Weierstrass

  3. 1872

    Weierstrass constructs a nowhere-differentiable continuous function, showing rigor is necessary

    Karl Weierstrass

Summary

  • lim_{x->a}f(x)=L means: for every epsilon>0 there exists delta>0 such that 0<|x-a|<delta implies |f(x)-L|<epsilon.
  • Intuition: epsilon-delta is a game -- for any target tolerance epsilon, you can find a neighborhood delta that keeps f(x) within epsilon of L.
  • Proof technique: start with |f(x)-L|<epsilon, work backwards to find delta, then write the forward proof.
  • Uniqueness: limits are unique (if they exist). The algebra of limits makes polynomials easy: lim p(x) = p(a).

References

  1. BookSpivak, M. Calculus. 4th ed. Publish or Perish, 2008.
  2. BookRudin, W. Principles of Mathematical Analysis. McGraw-Hill, 1976.