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

From Math Wiki
Jump to navigation Jump to search
Line 133: Line 133:
 
== Examples with Quadratic Functions ==
 
== Examples with Quadratic Functions ==
  
'''Problem 3.''' Using the definition of a limit, show that <math style="vertical-align: 0px">{\displaystyle \lim_{x\rightarrow1}x^{2}=1}</math>.
+
'''Problem 3.''' Using the definition of a limit, show that <math style="vertical-align: -13px">{\displaystyle \lim_{x\rightarrow1}x^{2}=1}</math>.
  
''Solution.'' Here, we have <math style="vertical-align: 0px">f(x)=x^{2},\ L=1</math> and <math style="vertical-align: 0px">c=1</math>. We again
+
''Solution.'' Here, we have <math style="vertical-align: -5px">f(x)=x^{2},\ L=1</math>&thinsp; and <math style="vertical-align: 0px">c=1</math>. We again
begin with scratchwork. Suppose <math style="vertical-align: 0px">|f(x)-L|<\epsilon</math>. We then solve
+
begin with scratchwork. Suppose <math style="vertical-align: -5px">|f(x)-L|<\epsilon</math>. We then solve
for <math style="vertical-align: 0px">|x-c|=|x-1|</math> to find  
+
for <math style="vertical-align: -5px">|x-c|=|x-1|</math> to find  
 +
 
 +
<br>
  
 
::<math>\begin{array}{ccccrcl}
 
::<math>\begin{array}{ccccrcl}
Line 148: Line 150:
 
\Rightarrow &  &  &  & |x-1| & < & {\displaystyle \frac{\epsilon}{|x+1|}.}
 
\Rightarrow &  &  &  & |x-1| & < & {\displaystyle \frac{\epsilon}{|x+1|}.}
 
\end{array}</math>  
 
\end{array}</math>  
 +
<br>
  
This certainly describes <math style="vertical-align: 0px">|x-1|</math> in terms of <math style="vertical-align: 0px">\epsilon</math>, but there's
+
This certainly describes <math style="vertical-align: -5px">|x-1|</math> in terms of <math style="vertical-align: 0px">\epsilon</math>, but there's
also an <math style="vertical-align: 0px">x</math> on the right hand side! This requires us to pick an ``initial''
+
also an <math style="vertical-align: 0px">x</math> on the right hand side! This requires us to pick an "initial"
</math>\delta</math>. Let's choose <math style="vertical-align: 0px">\delta=1</math>. Then, whenever <math style="vertical-align: 0px">|x-c|=|x-1|<\delta=1</math>,
+
<math style="vertical-align: 0px">\delta</math>. Let's choose <math style="vertical-align: -2px">\delta=1</math>. Then, whenever <math style="vertical-align: -5px">|x-c|=|x-1|<\delta=1</math>,
 
we have  
 
we have  
  
 
::<math>-\delta\,=\,-1\,<\, x-1\,<\,1\,=\,\delta,</math>
 
::<math>-\delta\,=\,-1\,<\, x-1\,<\,1\,=\,\delta,</math>
  
in the manner explained in '''An Explanation'. More importantly, by
+
in the manner explained in '''An Explanation'''. More importantly, by
 
adding two to the inequality we have  
 
adding two to the inequality we have  
  
Line 166: Line 169:
 
::<math>\frac{\epsilon}{3}\,<\,\frac{\epsilon}{x+1}\,=\,\frac{\epsilon}{|x+1|}\,<\,\frac{\epsilon}{1}\,=\,\epsilon.</math>  
 
::<math>\frac{\epsilon}{3}\,<\,\frac{\epsilon}{x+1}\,=\,\frac{\epsilon}{|x+1|}\,<\,\frac{\epsilon}{1}\,=\,\epsilon.</math>  
  
This means that for any <math style="vertical-align: 0px">x</math> satisfying <math style="vertical-align: 0px">|x-1|<1</math>, we know that <math style="vertical-align: 0px">\epsilon/3<\epsilon/|x+1|</math>.
+
This means that for any <math style="vertical-align: 0px">x</math> satisfying <math style="vertical-align: -5px">|x-1|<1</math>, we know that <math style="vertical-align: -5px">\epsilon/3<\epsilon/|x+1|</math>.
 
Thus, we can choose a <math style="vertical-align: 0px">\delta<\epsilon/3</math>, and the proof should work.
 
Thus, we can choose a <math style="vertical-align: 0px">\delta<\epsilon/3</math>, and the proof should work.
 
There's a small problem, though - we already chose a <math style="vertical-align: 0px">\delta=1</math>.
 
There's a small problem, though - we already chose a <math style="vertical-align: 0px">\delta=1</math>.
The way around this is to use the minimum function.
+
The way around this is to use the '''minimum function''':
  
::When we write <math style="vertical-align: 0px">\min\{a,b\}</math>, it means to take whichever is the
+
<blockquote>When we write <math style="vertical-align: -5px">\min\{a,b\}</math>, it means to take whichever is the
least of both <math style="vertical-align: 0px">a</math> and <math style="vertical-align: 0px">b</math>.  
+
least of both <math style="vertical-align: 0px">a</math> and <math style="vertical-align: 0px">b</math>. </blockquote>
  
 
Now, our proof can be written.
 
Now, our proof can be written.
  
'''Proof.''' Let <math style="vertical-align: 0px">\epsilon>0</math> be given. Choose <math style="vertical-align: 0px">\delta=\min\{1,\epsilon/3\}</math>.
+
'''Proof.''' Let <math style="vertical-align: 0px">\epsilon>0</math>&thinsp; be given. Choose <math style="vertical-align: -5px">\delta=\min\{1,\epsilon/3\}</math>.
Then, whenever <math style="vertical-align: 0px">|x-c|=|x-1|<\delta</math>,  
+
Then, whenever <math style="vertical-align: -5px">|x-c|=|x-1|<\delta</math>,  
 +
 
 +
<br>
  
 
::<math>\begin{array}{ccrcrclccccc}
 
::<math>\begin{array}{ccrcrclccccc}
Line 195: Line 200:
 
</math>\square</math>
 
</math>\square</math>
  
'''Problem 4.''' Using the definition of a limit, show that <math style="vertical-align: 0px">{\displaystyle \lim_{x\rightarrow2}3x^{2}=12}</math>.
+
'''Problem 4.''' Using the definition of a limit, show that <math style="vertical-align: -12px">{\displaystyle \lim_{x\rightarrow2}3x^{2}=12}</math>.
  
 
''Solution.'' This time, <math style="vertical-align: 0px">f(x)=3x^{2},\ L=12</math> and <math style="vertical-align: 0px">c=2</math>. We follow
 
''Solution.'' This time, <math style="vertical-align: 0px">f(x)=3x^{2},\ L=12</math> and <math style="vertical-align: 0px">c=2</math>. We follow
Line 246: Line 251:
 
</math>\square</math>
 
</math>\square</math>
  
'''Problem 5.'''Using the definition of a limit, show that <math style="vertical-align: 0px">{\displaystyle \lim_{x\rightarrow2}x^{2}+3x+1=11}</math>.  
+
'''Problem 5.'''Using the definition of a limit, show that <math style="vertical-align: -12px">{\displaystyle \lim_{x\rightarrow2}x^{2}+3x+1=11}</math>.  
  
 
''Solution.'' Here, we have <math style="vertical-align: 0px">f(x)=x^{2}+3x+1,\ L=11</math> and <math style="vertical-align: 0px">c=2</math>.
 
''Solution.'' Here, we have <math style="vertical-align: 0px">f(x)=x^{2}+3x+1,\ L=11</math> and <math style="vertical-align: 0px">c=2</math>.

Revision as of 12:38, 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

A few quick notes about these types of proofs:

  • 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.

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" . 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 and   to complete the proof.

There are many more difficult examples, but these are meant as an introduction.