ConceptComplete

Limits and Continuity - Core Definitions

The concept of a limit is the foundation upon which all of calculus is built. Understanding limits allows us to rigorously define continuity, derivatives, and integrals, making them essential to mathematical analysis.

DefinitionLimit of a Function

Let ff be a function defined on an open interval containing aa, except possibly at aa itself. We say that the limit of f(x)f(x) as xx approaches aa is LL, written as limxaf(x)=L\lim_{x \to a} f(x) = L if for every ϵ>0\epsilon > 0, there exists a δ>0\delta > 0 such that whenever 0<xa<δ0 < |x - a| < \delta, we have f(x)L<ϵ|f(x) - L| < \epsilon.

This epsilon-delta definition, formalized by Weierstrass, captures the intuitive notion that we can make f(x)f(x) arbitrarily close to LL by taking xx sufficiently close to aa. The condition 0<xa0 < |x - a| ensures that we never actually evaluate ff at aa itself, which is crucial since the limit may exist even when f(a)f(a) is undefined.

DefinitionOne-Sided Limits

We define right-hand limit as limxa+f(x)=L\lim_{x \to a^+} f(x) = L if for every ϵ>0\epsilon > 0, there exists δ>0\delta > 0 such that a<x<a+δa < x < a + \delta implies f(x)L<ϵ|f(x) - L| < \epsilon.

Similarly, the left-hand limit is limxaf(x)=L\lim_{x \to a^-} f(x) = L if for every ϵ>0\epsilon > 0, there exists δ>0\delta > 0 such that aδ<x<aa - \delta < x < a implies f(x)L<ϵ|f(x) - L| < \epsilon.

Remark

A limit limxaf(x)=L\lim_{x \to a} f(x) = L exists if and only if both one-sided limits exist and are equal: limxaf(x)=limxa+f(x)=L\lim_{x \to a^-} f(x) = \lim_{x \to a^+} f(x) = L This characterization is often useful for analyzing piecewise functions.

ExampleComputing a Simple Limit

Consider limx2(3x1)\lim_{x \to 2} (3x - 1). Intuitively, as xx approaches 2, 3x13x - 1 approaches 3(2)1=53(2) - 1 = 5. To prove this rigorously, let ϵ>0\epsilon > 0 be given. We need to find δ>0\delta > 0 such that 0<x2<δ    3x15<ϵ0 < |x - 2| < \delta \implies |3x - 1 - 5| < \epsilon

Simplifying: 3x6=3x2<ϵ|3x - 6| = 3|x - 2| < \epsilon, so x2<ϵ/3|x - 2| < \epsilon/3. Therefore, choosing δ=ϵ/3\delta = \epsilon/3 works, and we conclude limx2(3x1)=5\lim_{x \to 2} (3x - 1) = 5.

Understanding these foundational definitions is essential for developing the limit laws and continuity properties that follow.