Difference between revisions of "Limit of a Function(Definition): Introduction to ε-δ Arguments"

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When most students initially confront the definition above, they are
 
When most students initially confront the definition above, they are
 
really confused. It helps to ``unravel'' the absolute values a bit.
 
really confused. It helps to ``unravel'' the absolute values a bit.
For example, <math style="vertical-align: 0px">|f(x)-L|<\epsilon</math> is really the same thing as  
+
For example, <math style="vertical-align: -5px">|f(x)-L|<\epsilon</math> is really the same thing as  
  
 
::<math>-\epsilon\,<\, f(x)-L\,<\,\epsilon.</math>  
 
::<math>-\epsilon\,<\, f(x)-L\,<\,\epsilon.</math>  
Line 18: Line 18:
 
::<math>L-\epsilon\,<\, f(x)\,<\, L+\epsilon.</math>
 
::<math>L-\epsilon\,<\, f(x)\,<\, L+\epsilon.</math>
  
In other words, we are trying to restrict <math style="vertical-align: 0px">f(x)</math> to the''' <math style="vertical-align: 0px">\epsilon</math>-neighborhood
+
In other words, we are trying to restrict <math style="vertical-align: -4px">f(x)</math> to the '''<math style="vertical-align: 0px">\boldsymbol{\epsilon}</math>-neighborhood centered at''' <math style="vertical-align: 0px">\boldsymbol{L}</math>, which is the interval <math style="vertical-align: -5px">(L-\epsilon,L+\epsilon)</math>.
centered at <math style="vertical-align: 0px">L</math>''', which is the interval <math style="vertical-align: 0px">(L-\epsilon,L+\epsilon)</math>.
+
Similarly, we can rewrite <math style="vertical-align: -5px">|x-c|<\delta</math> to become
Similarly, we can rewrite <math style="vertical-align: 0px">|x-c|<\delta</math> to become
 
  
 
::<math>c-\delta\,<\, x\,<\, c+\delta.</math>  
 
::<math>c-\delta\,<\, x\,<\, c+\delta.</math>  
  
 
On its own, this would mean <math style="vertical-align: 0px">x</math> lies in the <math style="vertical-align: 0px">\delta</math>-neighborhood
 
On its own, this would mean <math style="vertical-align: 0px">x</math> lies in the <math style="vertical-align: 0px">\delta</math>-neighborhood
centered at <math style="vertical-align: 0px">c,</math> which is the interval <math style="vertical-align: 0px">(c-\delta,c+\delta)</math>. But
+
centered at <math style="vertical-align: 0px">c</math>, which is the interval <math style="vertical-align: -5px">(c-\delta,c+\delta)</math>. But
what about the other requirement, that <math style="vertical-align: 0px">0<|x-c|?</math> This means we ignore
+
what about the other requirement, that <math style="vertical-align: -5px">0<|x-c|</math>? This means we ignore
what happens at <math style="vertical-align: 0px">c</math>. This neighborhood - the interval <math style="vertical-align: 0px">(c-\delta,c+\delta)</math>,
+
what happens at <math style="vertical-align: 0px">c</math>.
 +
 
 +
This neighborhood - the interval <math style="vertical-align: -5px">(c-\delta,c+\delta)</math>,
 
minus the point <math style="vertical-align: 0px">c</math> - is known as a '''punctured neighborhood'''.  
 
minus the point <math style="vertical-align: 0px">c</math> - is known as a '''punctured neighborhood'''.  
  

Revision as of 11:58, 15 July 2015

Formal Definition

We say is the limit of at if, for any , there exists a   such that whenever ,

An Explanation

When most students initially confront the definition above, they are really confused. It helps to ``unravel the absolute values a bit. For example, is really the same thing as

If we then add to each term, we arrive at

In other words, we are trying to restrict to the -neighborhood centered at , which is the interval . Similarly, we can rewrite to become

On its own, this would mean lies in the -neighborhood centered at , which is the interval . But what about the other requirement, that ? This means we ignore what happens at .

This neighborhood - the interval , minus the point - is known as a punctured neighborhood.

This definition can possibly be better understood through a short video.

< Video coming soon! >

Proof Approach

The goal in these proofs is always the same: we need to find a , which will usually be expressed in terms of an arbitrary . For example, we might have to choose a , or a , or even a . But in each case, we usually build our proof ``backwards in what we can refer to as scratchwork.

Scratchwork begins by assuming our desired result, that

From here, we do whatever it takes (usually factoring) to change into . Once we have a statement of the form

this allows us to pick a that will work. We then just ``reverse the chain of equalities in our scratchwork to construct the proof. It's easier to see in a few examples.

Examples with Linear Functions

Problem 1. Using the definition of a limit, show that .

Solution. Looking at the statement we need to prove, we have and . Since for any , we have , we know that for any

as must be strictly positive. This means any will work. To write it out, you would proceed as follows:

Proof. Let be given. Choose . Then, whenever , we have

</math>\square</math>

A few quick notes about these.

Every one will begin with ``let be given. That's because most of the time, our will be tied to the .

We conclude the proof with a box/square, indicating we're done.

Problem 2. Using the definition of a limit, show that .

Solution. In this case, we have and . We again begin with scratchwork, and assume our goal. If we knew that , then we would work to get an expression that has . It goes like this:

This gives us our . We will choose a .

But that was all scratchwork, and the formal writeup looks like a bit different:

Proof. Let be given. Choose . Then, whenever , we have

as required.

</math>\square</math>

It should be mentioned that in cases where , and , we will get that every time.

Examples with Quadratic Functions

Problem 3. Using the definition of a limit, show that .

Solution. Here, we have and . We again begin with scratchwork. Suppose . We then solve for to find

This certainly describes in terms of , but there's also an on the right hand side! This requires us to pick an ``initial </math>\delta</math>. Let's choose . Then, whenever , we have

in the manner explained in An Explanation'. More importantly, by adding two to the inequality we have

Dividing by this inequality (which reverses its direction), we have

This means that for any satisfying , we know that . Thus, we can choose a , and the proof should work. There's a small problem, though - we already chose a . The way around this is to use the minimum function.

When we write , it means to take whichever is the

least of both and .

Now, our proof can be written.

Proof. Let be given. Choose . Then, whenever ,

as required.

</math>\square</math>

Problem 4. Using the definition of a limit, show that .

Solution. This time, and . We follow the same pattern, doing the scratchwork first. Assume . Then

Based on the previous problem, let's choose an initial . Then we have

Now adding 4 to the inequality, we have

Dividing by this inequality, we have

So we can choose Then the proof works.

Proof. Let be given. Choose . Then, whenever ,

as required.

</math>\square</math>

Problem 5.Using the definition of a limit, show that .

Solution. Here, we have and . We again begin with scratchwork, assuming . Then

Now, we again assume , so

and adding 7 to the inequality,

Again, dividing by we have

for all satisfying We can now write the proof.

Proof. Let be given. Choose . Then, whenever

as required.

</math>\square</math>

Other Examples

Problem 6. Using the definition of a limit, show that .

Solution. We have and . We start our scratchwork as usual, assuming . Then we have

Similar to the quadratic examples, we can set an initial Then

and adding one to each term we find

This, in turn, means that

Multiplying the inequality by , we have

for all satisfying . This gives us our , and we can write the proof.

Proof. Let be given. Choose . Then, whenever ,

as required. Notice that we used both of the inequalities involving </math>\epsilon</math> and to complete the proof.

</math>\square</math>

There are many more difficult examples, but these are meant as an introduction. \end{document}