The Precise Definition of a Limit: Learn It 2

Advanced Applications of the Epsilon-Delta Definition: Proofs, Non-Existence, and Algebraic Calculations

Using the Epsilon-Delta Definition of Limits

We will demonstrate how to use the epsilon-delta definition of a limit to construct a rigorous proof of one of the limit laws. The triangle inequality is a key component of this proof, so let’s review it first.

triangle inequality

The triangle inequality states that if aa and bb are any real numbers, then |a+b||a|+|b||a+b||a|+|b|.

Proof


We prove the following limit law: If limxaf(x)=Llimxaf(x)=L and limxag(x)=Mlimxag(x)=M, then limxa(f(x)+g(x))=L+Mlimxa(f(x)+g(x))=L+M.

Let ε>0ε>0.

Choose δ1>0δ1>0 so that if 0<|xa|<δ10<|xa|<δ1, then |f(x)L|<ε/2|f(x)L|<ε/2.

Choose δ2>0δ2>0 so that if 0<|xa|<δ20<|xa|<δ2, then |g(x)M|<ε/2|g(x)M|<ε/2.

Choose δ=min{δ1,δ2}δ=min{δ1,δ2}.

Assume 0<|xa|<δ0<|xa|<δ.

Thus,

0<|xa|<δ10<|xa|<δ1 and 0<|xa|<δ20<|xa|<δ2

Hence,

|(f(x)+g(x))(L+M)|=|(f(x)L)+(g(x)M)||f(x)L|+|g(x)M|<ε2+ε2=ε|(f(x)+g(x))(L+M)|=|(f(x)L)+(g(x)M)||f(x)L|+|g(x)M|<ε2+ε2=ε

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Exploring the Non-Existence of Limits

We now explore what it means for a limit not to exist. The limit limxaf(x)limxaf(x) does not exist if there is no real number LL for which limxaf(x)=Llimxaf(x)=L. For all real numbers LL, limxaf(x)Llimxaf(x)L.

To understand what this means, we look at each part of the definition of limxaf(x)=Llimxaf(x)=L together with its opposite. 

Translation of the Definition of limxaf(x)=Llimxaf(x)=L and its Opposite
Definition Opposite
1. For every ε>0ε>0, 1. There exists ε>0ε>0 so that
2. there exists a δ>0δ>0 so that 2. for every δ>0δ>0,
3. if 0<|xa|<δ0<|xa|<δ, then |f(x)L|<ε|f(x)L|<ε. 3. There is an xx satisfying 0<|xa|<δ0<|xa|<δ so that |f(x)L|ε.

Finally, we may state what it means for a limit not to exist. The limit limxaf(x) does not exist if for every real number L, there exists a real number ε>0 so that for all δ>0, there is an x satisfying 0<|xa|<δ, so that |f(x)L|ε.

Let’s apply this in the example to show that a limit does not exist.

Show that limx0|x|x does not exist. The graph of f(x)=|x|x is shown here:

"A

Finding Deltas Algebraically for Given Epsilons

Now that we have proven limits, we can now apply them with actual numbers for ε and δ. Think of ε as the error in the x-direction and δ to be the error in the y-direction. These have applications in engineering when these errors are considered tolerances. We want to know what the error intervals are, and we are trying to minimize these errors.

Find an open interval about x0 on which the inequality |f(x)L|<0 holds. Then give the largest value δ>0 such that for all x satisfying 0<|xx0|<δ the inequality |f(x)L|<ε holds.

f(x)=2x8,L=6,x0=7,ε=0.14