View source for Limit of a Function(Definition): Introduction to ε-δ Arguments

== Formal Definition ==

We say <math style="vertical-align: 0px">L</math> is the limit of <math style="vertical-align: -5px">f</math> at <math style="vertical-align: 0px">c</math> if, for any <math style="vertical-align: 0px">\epsilon>0</math>,
there exists a <math style="vertical-align: 0px">\delta>0</math>&thinsp; such that whenever <math style="vertical-align: -5px">0<|x-c|<\delta</math>,

::<math>\left|f(x)-L\right|\,<\,\epsilon.</math>

== 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, <math style="vertical-align: -5px">|f(x)-L|<\epsilon</math> is really the same thing as 

::<math>-\epsilon\,<\, f(x)-L\,<\,\epsilon.</math> 

If we then add <math style="vertical-align: 0px">L</math> to each term, we arrive at 

::<math>L-\epsilon\,<\, f(x)\,<\, L+\epsilon.</math>

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>.
Similarly, we can rewrite <math style="vertical-align: -5px">|x-c|<\delta</math> to become

::<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
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: -5px">0<|x-c|</math>&thinsp;? This means we ignore
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'''. 

This definition can possibly be better understood through a short
video (best viewed fullscreen).
<br><br>

<div style="text-align:center">'''''< Does not work on Internet Explorer >'''''</div>

<center><HTML5video width="640" height="420" autoplay="false">LimitDef</HTML5video></center>

<br>

== Proof Approach ==

The goal in these proofs is always the same: we need to find a <math style="vertical-align: 0px">\delta</math>,
which will usually be expressed in terms of an arbitrary <math style="vertical-align: 0px">\epsilon</math>.
For example, we might have to choose a <math style="vertical-align: 0px">\delta<\epsilon</math>, or a <math style="vertical-align: -5px">\delta<\epsilon/3</math>,
or even a <math style="vertical-align: -5px">\delta<\min\{1,\epsilon/3\}</math>. But in each case, we usually
build our proof <u>''backwards''</u> in what we can refer to as scratchwork. 

Scratchwork begins by assuming our desired result, that 

::<math>|f(x)-L|\,<\,\epsilon.</math> 

From here, we do whatever it takes (usually factoring) to change <math style="vertical-align: -5px">|f(x)-L|</math>
into <math style="vertical-align: -5px">|x-c|</math>. Once we have a statement of the form 

::<math>|x-c|\,<\,\textrm{something},</math>

this allows us to pick a <math style="vertical-align: 0px">\delta</math> that will work. We then just <u>''reverse''</u>
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 <math style="vertical-align: -12px">{\displaystyle \lim_{x\rightarrow1}10=10}</math>.

''Solution.'' Looking at the statement we need to prove, we have <math style="vertical-align: -5px">f(x)=10,\ L=10</math>
&thinsp;and <math style="vertical-align: -1px">c=1</math>. Since <math style="vertical-align: -5px">f(x)=10</math> for all <math style="vertical-align:0px">x</math>, we know that for any <math style="vertical-align: -4px">x,</math> 

::<math>|f(x)-L|\,=\,|10-10|\,=\,0\,<\epsilon</math> 

as <math style="vertical-align: 0px">\epsilon</math> must be strictly positive. This means any <math style="vertical-align: 0px">\delta</math>
will work. To write it out formally, you would proceed as follows:

'''Proof.''' Let <math style="vertical-align: 0px">\epsilon>0</math>&thinsp; be given. Choose <math style="vertical-align: -1px">\delta=1</math>. Then,
whenever <math style="vertical-align: -5px">0<|x-c|=|x-1|<\delta=1</math>, we have 

::<math>|f(x)-L|\ =\ |10-10|\ =\ 0\ <\ \epsilon.</math>

<div style="text-align:right"><math>\square</math></div>
A few quick notes about these types of proofs:

*Every one will begin with "let <math style="vertical-align: 0px">\epsilon>0</math>&thinsp; be given." That's because most of the time, our <math style="vertical-align: 0px">\delta</math> will be tied to the <math style="vertical-align: 0px">\epsilon</math>. 

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

<br>

'''Problem 2.''' Using the definition of a limit, show that <math style="vertical-align: -12px">{\displaystyle \lim_{x\rightarrow3}2x-4=2}</math>.

''Solution.'' In this case, we have <math style="vertical-align: -5px">f(x)=2x-4,\ L=2</math>&thinsp; and <math style="vertical-align: 0px">c=3</math>.
We again begin with scratchwork, and assume our goal. If we <u>''knew''</u>
that <math style="vertical-align: -5px">|f(x)-L|<\epsilon</math>, then we would work to get an expression
that has <math style="vertical-align: -5px">|x-c|=|x-3|</math>. It goes like this:

<br>

::<math>\begin{array}{ccccrcl}
 &  &  &  & \left|(2x-4)-2\right| & < & \epsilon\\
\\
\Rightarrow &  &  &  & |2x-6| & < & \epsilon\\
\\
\Rightarrow &  &  &  & |2(x-3)| & < & \epsilon\\
\\
\Rightarrow &  &  &  & 2|x-3| & < & \epsilon\\
\\
\Rightarrow &  &  &  & |x-3| & < & {\displaystyle \frac{\epsilon}{2}.}
\end{array}</math> 
<br>

This gives us our <math style="vertical-align: 0px">\delta</math>. We will choose a <math style="vertical-align: -5px">\delta<\epsilon/2</math>. 

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

'''Proof.''' Let <math style="vertical-align: 0px">\epsilon>0</math>&thinsp; be given. Choose <math style="vertical-align: -5px">\delta<\epsilon/2</math>.
Then, whenever <math style="vertical-align: -5px">0<|x-c|=|x-3|<\delta<\epsilon/2</math>, we have 

<br>

::<math>\begin{array}{ccccrcl}
 &  &  &  & |x-3| & < & {\displaystyle \frac{\epsilon}{2}}\\
\\
\Rightarrow &  &  &  & |2x-6| & < & \epsilon\\
\\
\Rightarrow &  &  &  & |2(x-3)| & < & \epsilon\\
\\
\Rightarrow &  &  &  & |2x-6| & < & \epsilon\\
\\
\Rightarrow &  &  &  & |(2x-4)-2| & < & {\displaystyle \epsilon}\\
\\
\Rightarrow &  &  &  & |f(x)-L| & < & \epsilon,
\end{array}</math>
<br>

as required. 

<div style="text-align:right;"><math>\square</math></div>

It should be mentioned that in cases where <math style="vertical-align: -5px">f(x)=mx+b</math>, and <math style="vertical-align: -5px">m\neq0</math>,
we will get that <math style="vertical-align: -5px">\delta<\epsilon/|m|</math> every time.

== Examples with Quadratic Functions ==

'''Problem 3.''' Using the definition of a limit, show that <math style="vertical-align: -12px">{\displaystyle \lim_{x\rightarrow1}x^{2}=1}</math>.

''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: -5px">|f(x)-L|<\epsilon</math>. We then solve
for <math style="vertical-align: -5px">|x-c|=|x-1|</math> to find 

<br>

::<math>\begin{array}{ccccrcl}
 &  &  &  & \left|x^{2}-1\right| & < & \epsilon\\
\\
\Rightarrow &  &  &  & |(x-1)(x+1)| & < & \epsilon\\
\\
\Rightarrow &  &  &  & |x+1|\cdot|x-1| & < & \epsilon\\
\\
\Rightarrow &  &  &  & |x-1| & < & {\displaystyle \frac{\epsilon}{|x+1|}.}
\end{array}</math> 
<br>

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"
<math style="vertical-align: 0px">\delta</math>. Let's choose <math style="vertical-align: -1px">\delta=1</math>. Then, whenever <math style="vertical-align: -5px">0<|x-c|=|x-1|<\delta=1</math>,
we have 

::<math>-\delta\,=\,-1\,<\, x-1\,<\,1\,=\,\delta,</math>

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

::<math>1\,<\, x+1\,<\,3.\qquad\qquad\qquad(\dagger)</math> 

Dividing <math style="vertical-align: 0px">\epsilon</math> by this inequality (which reverses its direction),
we have 

::<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: -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: -5px">\delta<\epsilon/3</math>, and the proof should work.
There's a small problem, though - we already chose a <math style="vertical-align: -1px">\delta=1</math>.
The way around this is to use the '''minimum function''':

<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>. </blockquote>

Now, our proof can be written.

'''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: -5px">0<|x-c|=|x-1|<\delta</math>, 

<br>

::<math>\begin{array}{ccrcrclccccc}
 &  &  &  & |x-1| & < & {\displaystyle \frac{\epsilon}{3}}\\
\\
\Rightarrow &  &  &  & 3|x-1| & < & \epsilon\\
\\
\Rightarrow &  & |x+1||x-1| & < & 3|x-1| & < & \epsilon &  &  &  &  & \textrm{(using~}\dagger)\\
\\
\Rightarrow &  &  &  & |x^{2}-1| & < & \epsilon\\
\\
\Rightarrow &  &  &  & |f(x)-L| & < & \epsilon,
\end{array}</math>

<br>
as required. 

<div style="text-align:right;"><math>\square</math></div>

<br>

'''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: -5px">f(x)=3x^{2},\ L=12</math>&thinsp; and <math style="vertical-align: 0px">c=2</math>. We follow
the same pattern, doing the scratchwork first. Assume <math style="vertical-align: -5px">|f(x)-L|<\epsilon</math>.
Then

<br>

::<math>\begin{array}{ccccrcl}
 &  &  &  & \left|3x^{2}-12\right| & < & \epsilon\\
\\
\Rightarrow &  &  &  & 3|x^{2}-4| & < & \epsilon\\
\\
\Rightarrow &  &  &  & 3|(x+2|\cdot|x-2)| & < & \epsilon\\
\\
\Rightarrow &  &  &  & |x-2| & < & {\displaystyle \frac{\epsilon}{3|x+2|}.}
\end{array}</math> 

<br>

Based on the previous problem, let's choose an initial <math style="vertical-align: -1px">\delta=1</math>.
Then we have 

::<math>-\delta\,=\,-1\,<\, x-2\,<\,1\,=\,\delta.</math>

Now adding 4 to the inequality, we have

::<math>3\,<\, x+2\,<\,5.\qquad\qquad\qquad(\dagger\dagger)</math> 

Dividing <math style="vertical-align: 0px">\epsilon</math> by this inequality, we have 

::<math>\frac{\epsilon}{5}\,<\,\frac{\epsilon}{|x+2|}\,<\,\frac{\epsilon}{3}.</math>

So we can choose <math style="vertical-align: -15px">\delta=\min\left\{ 1,\frac{1}{3}\cdot\frac{\epsilon}{5}\right\} =\min\left\{ 1,\frac{\epsilon}{15}\right\}.</math> 
Then the proof works.

'''Proof.''' Let <math style="vertical-align: 0px">\epsilon>0</math>&thinsp; be given. Choose <math style="vertical-align: -16px">\delta=\min\left\{ 1,\frac{\epsilon}{15}\right\}</math>.
Then, whenever <math style="vertical-align: -5px">0<|x-c|=|x-2|<\delta</math>, 

<br>

::<math>\begin{array}{ccrcrclccccc}
 &  &  &  & |x-2| & < & {\displaystyle \frac{\epsilon}{15}}\\
\\
\Rightarrow &  &  &  & 15|x-2| & < & \epsilon\\
\\
\Rightarrow &  & 3|x+2||x-2| & < & 15|x-2| & < & \epsilon &  &  &  &  & \textrm{(using~}\dagger\dagger)\\
\\
\Rightarrow &  &   3|x^{2}-4|& & & < & \epsilon\\
\\
\Rightarrow &  &   |3x^{2}-12|& & & < & \epsilon\\
\\
\Rightarrow &  &  &  & |f(x)-L| & < & \epsilon,
\end{array}</math>
<br>

as required.

<div style="text-align:right;"><math>\square</math></div>

<br>

'''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: -5px">f(x)=x^{2}+3x+1,\ L=11</math>&thinsp; and <math style="vertical-align: 0px">c=2</math>.
We again begin with scratchwork, assuming <math style="vertical-align: -5px">|f(x)-L|<\epsilon</math>. Then

<br>

::<math>\begin{array}{ccccrcl}
 &  &  &  & \left|x^{2}+3x+1-11\right| & < & \epsilon\\
\\
\Rightarrow &  &  &  & |x^{2}+3x-10| & < & \epsilon\\
\\
\Rightarrow &  &  &  & |(x+5)(x-2)| & < & \epsilon\\
\\
\Rightarrow &  &  &  & |(x+5|\cdot|x-2)| & < & \epsilon\\
\\
\Rightarrow &  &  &  & |x-2| & < & {\displaystyle \frac{\epsilon}{|x+5|}.}
\end{array}</math>

Now, we again assume <math style="vertical-align: -1px">\delta=1</math>, so 

::<math>-\delta\,=\,-1\,<\, x-2\,<\,1\,=\,\delta,</math>

and adding seven to the inequality,

::<math>-\delta\,=\,6\,<\, x+5\,<\,1\,=\,7.\qquad\qquad\qquad(\natural)</math>

Again, dividing <math style="vertical-align: 0px">\epsilon</math> by we have

::<math>\frac{\epsilon}{7}\,<\,\frac{\epsilon}{|x+5|}\,<\,\frac{\epsilon}{5}</math> 

for all <math style="vertical-align: 0px">x</math> satisfying <math style="vertical-align: -5px">|x-2|<1.</math> We can now write the proof.

'''Proof.''' Let <math style="vertical-align: 0px">\epsilon>0</math>&thinsp; be given. Choose <math style="vertical-align: -16px">\delta=\min\left\{ 1,\frac{\epsilon}{7}\right\}</math>. 
Then, whenever <math style="vertical-align: -5px">0<|x-c|=|x-2|<\delta,</math> 

::<math>\begin{array}{ccrcrclccccl}
 &  &  &  & |x-2| & < & {\displaystyle \frac{\epsilon}{7}}\\
\\
\Rightarrow &  &  &  & 7|x-2| & < & \epsilon\\
\\
\Rightarrow &  & |x+5||x-2| & < & 7|x-2| & < & \epsilon &  &  &  &  & \textrm{(using~}\natural)\\
\\
\Rightarrow &  & |x^{2}+3x-10| & & & < & \epsilon\\
\\
\Rightarrow &  &  |x^{2}+3x+1-11|& & & < & \epsilon\\
\\
\Rightarrow &  &  &  & |f(x)-L| & < & \epsilon,
\end{array}</math>
<br>

as required.

<div style="text-align:right;"><math>\square</math></div>

== Other Examples ==

'''Problem 6.''' Using the definition of a limit, show that <math style="vertical-align: -12px">{\displaystyle \lim_{x\rightarrow1}\sqrt{x}=1}</math>. 

''Solution.'' We have <math style="vertical-align: -5px">f(x)=\sqrt{x},\ L=1</math>&thinsp; and <math style="vertical-align: -1px">c=1</math>. We start
our scratchwork as usual, assuming <math style="vertical-align: -5px">|f(x)-L|<\epsilon</math>. Then we have 
<br><br>

::<math>\begin{array}{cccrcl}
 &  &  & |\sqrt{x}-1| & < & {\displaystyle \epsilon}\\
\\
\Rightarrow &  &  & |\sqrt{x}+1|\cdot|\sqrt{x}-1 & < & |\sqrt{x}+1|\cdot\epsilon\\
\\
\Rightarrow &  &  & |x-1| & < & |\sqrt{x}+1|\cdot\epsilon.
\end{array}</math>
<br>

Similar to the quadratic examples, we can set an initial <math style="vertical-align: -1px">\delta=1.</math>
Then

::<math>-\delta\,=\,-1\,<\, x-1\,<\,1\,=\,\delta,</math>

and adding one to each term we find

::<math>0\,<\, x\,<\,2.</math> 

This, in turn, means that 

::<math>1\,<\,\sqrt{x}+1\,<\,\sqrt{2}+1.</math> 

Multiplying the inequality by <math style="vertical-align: 0px">\epsilon</math>, we have 

::<math>\epsilon\,<\,|\sqrt{x}+1|\,<\,\left(\sqrt{2}+1\right)\epsilon\qquad\mbox{while}\qquad\frac{\epsilon}{\sqrt{2}+1}\,<\,\frac{\epsilon}{|\sqrt{x}+1|}\,<\,\frac{\epsilon}{1}.</math> 

for all <math style="vertical-align: 0px">x</math> satisfying <math style="vertical-align: -5px">|x-1|<1=\delta</math>. This gives us our <math style="vertical-align: 0px">\delta</math>,
and we can write the proof.

'''Proof.''' Let <math style="vertical-align: 0px">\epsilon>0</math>&thinsp; be given. Choose <math style="vertical-align: -5px">\delta=\min\left\{ 1,\epsilon\right\}</math>. 
Then, whenever <math style="vertical-align: -5px">|x-c|=|x-1|<\delta</math>,

<br>

::<math>\begin{array}{cccrclcc}
 &  &  & |x-1| & < & {\displaystyle \epsilon}\\
\\
\Rightarrow &  &  & |(\sqrt{x}+1)(\sqrt{x}-1)| & < & \epsilon\\
\\
\Rightarrow &  &  & \left|\sqrt{x}+1\right|\cdot\left|\sqrt{x}-1\right| & < & \epsilon\\
\\
\Rightarrow &  &  & \left|\sqrt{x}-1\right| & < & \displaystyle{\frac{\epsilon}{\left|\sqrt{x}+1\right|}} & < & \epsilon,
\end{array}</math>
<br>

as required. Notice that we used both of the inequalities involving <math style="vertical-align: 0px">\epsilon</math> and <math style="vertical-align: -10px">\left|\sqrt{x}+1\right|</math>&thinsp; to complete the proof.

<div style="text-align:right;"><math>\square</math></div>

<div style="text-align:center"> ''There are many more difficult examples, but these are meant as an introduction.''</div>