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Epsilon-Delta Definition of a Limit Worked Examples cheat sheet - grade college

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The epsilon-delta definition of a limit is the formal way calculus explains what it means for a function to approach a value. This cheat sheet helps students turn the definition into a clear proof strategy instead of a memorized sentence. It focuses on worked-example patterns for linear, quadratic, rational, and absolute value limits.

Students need this reference because epsilon-delta proofs require both algebraic control and precise logical wording.

The central idea is to make f(x)L<ε|f(x)-L|<\varepsilon happen by requiring 0<xa<δ0<|x-a|<\delta. Most examples work by rewriting or bounding f(x)L|f(x)-L| in terms of xa|x-a|. Once a useful inequality is found, choose δ\delta as a function of ε\varepsilon, often using a minimum such as δ=min{1,ε5}\delta=\min\{1,\frac{\varepsilon}{5}\}.

A complete proof states the choice of δ\delta, assumes 0<xa<δ0<|x-a|<\delta, and then shows f(x)L<ε|f(x)-L|<\varepsilon.

Key Facts

  • The formal definition is limxaf(x)=L\lim_{x\to a} f(x)=L if for every ε>0\varepsilon>0 there exists δ>0\delta>0 such that 0<xa<δ0<|x-a|<\delta implies f(x)L<ε|f(x)-L|<\varepsilon.
  • For a linear function f(x)=mx+bf(x)=mx+b, the limit proof often uses f(x)L=mxa|f(x)-L|=|m||x-a|, so choosing δ=εm\delta=\frac{\varepsilon}{|m|} works when m0m\ne0.
  • For limxax2=a2\lim_{x\to a} x^2=a^2, use x2a2=xax+a|x^2-a^2|=|x-a||x+a| and bound x+a|x+a| by first forcing xa<1|x-a|<1.
  • A common quadratic choice is δ=min{1,ε2a+1}\delta=\min\{1,\frac{\varepsilon}{2|a|+1}\} because x+a<2a+1|x+a|<2|a|+1 when xa<1|x-a|<1.
  • For a rational function, factor or combine fractions first, then use a restriction such as xa<1|x-a|<1 to keep denominators away from 00.
  • The condition 0<xa0<|x-a| means xx may approach aa but does not have to equal aa, so the function value f(a)f(a) may be undefined.
  • The number δ\delta is allowed to depend on ε\varepsilon, but it must be positive and chosen before assuming 0<xa<δ0<|x-a|<\delta.
  • Using δ=min{c,g(ε)}\delta=\min\{c,g(\varepsilon)\} lets one condition control nearby behavior while another condition guarantees f(x)L<ε|f(x)-L|<\varepsilon.

Vocabulary

Epsilon
Epsilon, written ε\varepsilon, is a positive tolerance for how close f(x)f(x) must be to the limit value LL.
Delta
Delta, written δ\delta, is a positive distance from aa that controls how close xx must be to aa.
Limit
The limit limxaf(x)=L\lim_{x\to a} f(x)=L means f(x)f(x) can be made arbitrarily close to LL by taking xx sufficiently close to aa.
Punctured neighborhood
A punctured neighborhood of aa is the set of points satisfying 0<xa<δ0<|x-a|<\delta.
Bounding
Bounding is the process of replacing a difficult factor with a simpler upper estimate, such as using x+a<2a+1|x+a|<2|a|+1.
Minimum choice
A minimum choice such as δ=min{1,ε5}\delta=\min\{1,\frac{\varepsilon}{5}\} enforces multiple inequalities at the same time.

Common Mistakes to Avoid

  • Choosing δ\delta before analyzing f(x)L|f(x)-L| is wrong because the proof must show exactly how closeness in xx forces closeness in f(x)f(x).
  • Forgetting the condition 0<xa0<|x-a| is wrong because the epsilon-delta definition concerns values near aa, not necessarily the value at aa.
  • Using δ=ε\delta=\varepsilon in every problem is wrong because nonlinear expressions such as x2a2|x^2-a^2| usually require extra bounds.
  • Dividing by a quantity that might be 00 is wrong because rational limit proofs must first restrict xx so the denominator stays safely away from 00.
  • Proving only one numerical case such as ε=0.01\varepsilon=0.01 is wrong because the definition requires the argument to work for every ε>0\varepsilon>0.

Practice Questions

  1. 1 Prove using the epsilon-delta definition that limx3(2x1)=5\lim_{x\to 3}(2x-1)=5, and give an explicit formula for δ\delta in terms of ε\varepsilon.
  2. 2 Prove that limx2x2=4\lim_{x\to 2}x^2=4 by bounding x+2|x+2| after assuming x2<1|x-2|<1.
  3. 3 Find a valid δ\delta choice for proving limx1x21x1=2\lim_{x\to 1}\frac{x^2-1}{x-1}=2 for x1x\ne1.
  4. 4 Explain why the value of f(a)f(a) does not affect whether limxaf(x)=L\lim_{x\to a}f(x)=L exists under the epsilon-delta definition.