# Limit of a Function(Definition): Introduction to ε-δ Arguments

## Formal Definition

We say ${\displaystyle L}$ is the limit of ${\displaystyle f}$ at ${\displaystyle c}$ if, for any ${\displaystyle \epsilon >0}$, there exists a ${\displaystyle \delta >0}$  such that whenever ${\displaystyle 0<|x-c|<\delta }$,

${\displaystyle \left|f(x)-L\right|\,<\,\epsilon .}$

## 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, ${\displaystyle |f(x)-L|<\epsilon }$ is really the same thing as

${\displaystyle -\epsilon \,<\,f(x)-L\,<\,\epsilon .}$

If we then add ${\displaystyle L}$ to each term, we arrive at

${\displaystyle L-\epsilon \,<\,f(x)\,<\,L+\epsilon .}$

In other words, we are trying to restrict ${\displaystyle f(x)}$ to the ${\displaystyle {\boldsymbol {\epsilon }}}$-neighborhood centered at ${\displaystyle {\boldsymbol {L}}}$, which is the interval ${\displaystyle (L-\epsilon ,L+\epsilon )}$. Similarly, we can rewrite ${\displaystyle |x-c|<\delta }$ to become

${\displaystyle c-\delta \,<\,x\,<\,c+\delta .}$

On its own, this would mean ${\displaystyle x}$ lies in the ${\displaystyle \delta }$-neighborhood centered at ${\displaystyle c}$, which is the interval ${\displaystyle (c-\delta ,c+\delta )}$. But what about the other requirement, that ${\displaystyle 0<|x-c|}$? This means we ignore what happens at ${\displaystyle c}$.

This neighborhood - the interval ${\displaystyle (c-\delta ,c+\delta )}$, minus the point ${\displaystyle c}$ - 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 ${\displaystyle \delta }$, which will usually be expressed in terms of an arbitrary ${\displaystyle \epsilon }$. For example, we might have to choose a ${\displaystyle \delta <\epsilon }$, or a ${\displaystyle \delta <\epsilon /3}$, or even a ${\displaystyle \delta <\min\{1,\epsilon /3\}}$. 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

${\displaystyle |f(x)-L|\,<\,\epsilon .}$

From here, we do whatever it takes (usually factoring) to change ${\displaystyle |f(x)-L|}$ into ${\displaystyle |x-c|}$. Once we have a statement of the form

${\displaystyle |x-c|\,<\,{\textrm {something}},}$

this allows us to pick a ${\displaystyle \delta }$ 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 ${\displaystyle {\displaystyle \lim _{x\rightarrow 1}10=10}}$.

Solution. Looking at the statement we need to prove, we have ${\displaystyle f(x)=10,\ L=10}$  and ${\displaystyle c=1}$. Since for any ${\displaystyle x}$, we have ${\displaystyle f(x)=10}$, we know that for any ${\displaystyle x}$,

${\displaystyle |f(x)-L|\,=\,|10-10|\,=\,0\,<\epsilon }$

as ${\displaystyle \epsilon }$ must be strictly positive. This means any ${\displaystyle \delta }$ will work. To write it out, you would proceed as follows:

Proof. Let ${\displaystyle \epsilon >0}$ be given. Choose ${\displaystyle \delta =1}$. Then, whenever ${\displaystyle |x-1|<\delta =1}$, we have

${\displaystyle |f(x)-L|\ =\ |10-10|\ =\ 0\ <\ \epsilon .}$
${\displaystyle \square }$

A few quick notes about these types of proofs:

• Every one will begin with "let ${\displaystyle \epsilon >0}$ be given." That's because most of the time, our ${\displaystyle \delta }$ will be tied to the ${\displaystyle \epsilon }$.
• We conclude the proof with a box/square, indicating we're done.

Problem 2. Using the definition of a limit, show that ${\displaystyle {\displaystyle \lim _{x\rightarrow 3}2x-4=2}}$.

Solution. In this case, we have ${\displaystyle f(x)=2x-4,\ L=2}$  and ${\displaystyle c=3}$. We again begin with scratchwork, and assume our goal. If we knew that ${\displaystyle |f(x)-L|<\epsilon }$, then we would work to get an expression that has ${\displaystyle |x-c|=|x-3|}$. It goes like this:

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

This gives us our ${\displaystyle \delta }$. We will choose a ${\displaystyle \delta <\epsilon /2}$.

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

Proof. Let ${\displaystyle \epsilon >0}$ be given. Choose ${\displaystyle \delta <\epsilon /2}$. Then, whenever ${\displaystyle |x-c|=|x-3|<\delta <\epsilon /2}$, we have

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

as required.

${\displaystyle \square }$

It should be mentioned that in cases where ${\displaystyle f(x)=mx+b}$, and ${\displaystyle m\neq 0}$, we will get that ${\displaystyle \delta <\epsilon /|m|}$ every time.

Problem 3. Using the definition of a limit, show that ${\displaystyle {\displaystyle \lim _{x\rightarrow 1}x^{2}=1}}$.

Solution. Here, we have ${\displaystyle f(x)=x^{2},\ L=1}$  and ${\displaystyle c=1}$. We again begin with scratchwork. Suppose ${\displaystyle |f(x)-L|<\epsilon }$. We then solve for ${\displaystyle |x-c|=|x-1|}$ to find

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

This certainly describes ${\displaystyle |x-1|}$ in terms of ${\displaystyle \epsilon }$, but there's also an ${\displaystyle x}$ on the right hand side! This requires us to pick an "initial" ${\displaystyle \delta }$. Let's choose ${\displaystyle \delta =1}$. Then, whenever ${\displaystyle |x-c|=|x-1|<\delta =1}$, we have

${\displaystyle -\delta \,=\,-1\,<\,x-1\,<\,1\,=\,\delta ,}$

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

${\displaystyle 1\,<\,x+1\,<\,3.\qquad \qquad \qquad (\dagger )}$

Dividing ${\displaystyle \epsilon }$ by this inequality (which reverses its direction), we have

${\displaystyle {\frac {\epsilon }{3}}\,<\,{\frac {\epsilon }{x+1}}\,=\,{\frac {\epsilon }{|x+1|}}\,<\,{\frac {\epsilon }{1}}\,=\,\epsilon .}$

This means that for any ${\displaystyle x}$ satisfying ${\displaystyle |x-1|<1}$, we know that ${\displaystyle \epsilon /3<\epsilon /|x+1|}$. Thus, we can choose a ${\displaystyle \delta <\epsilon /3}$, and the proof should work. There's a small problem, though - we already chose a ${\displaystyle \delta =1}$. The way around this is to use the minimum function:

When we write ${\displaystyle \min\{a,b\}}$, it means to take whichever is the least of both ${\displaystyle a}$ and ${\displaystyle b}$.

Now, our proof can be written.

Proof. Let ${\displaystyle \epsilon >0}$  be given. Choose ${\displaystyle \delta =\min\{1,\epsilon /3\}}$. Then, whenever ${\displaystyle |x-c|=|x-1|<\delta }$,

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

as required.

${\displaystyle \square }$

Problem 4. Using the definition of a limit, show that ${\displaystyle {\displaystyle \lim _{x\rightarrow 2}3x^{2}=12}}$.

Solution. This time, ${\displaystyle f(x)=3x^{2},\ L=12}$  and ${\displaystyle c=2}$. We follow the same pattern, doing the scratchwork first. Assume ${\displaystyle |f(x)-L|<\epsilon }$. Then

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

Based on the previous problem, let's choose an initial ${\displaystyle \delta =1}$. Then we have

${\displaystyle -\delta \,=\,-1\,<\,x-2\,<\,1\,=\,\delta .}$

Now adding 4 to the inequality, we have

${\displaystyle 3\,<\,x+2\,<\,5.\qquad \qquad \qquad (\dagger \dagger )}$

Dividing ${\displaystyle \epsilon }$ by this inequality, we have

${\displaystyle {\frac {\epsilon }{5}}\,<\,{\frac {\epsilon }{|x+2|}}\,<\,{\frac {\epsilon }{3}}.}$

So we can choose ${\displaystyle \delta =\min \left\{1,{\frac {1}{3}}\cdot {\frac {\epsilon }{3}}\right\}=\min \left\{1,{\frac {\epsilon }{9}}\right\}.}$ Then the proof works.

Proof. Let ${\displaystyle \epsilon >0}$  be given. Choose ${\displaystyle \delta =\min \left\{1,{\frac {\epsilon }{9}}\right\}}$. Then, whenever ${\displaystyle |x-c|=|x-2|<\delta }$,

${\displaystyle {\begin{array}{ccrcrclccccc}&&&&|x-2|&<&{\displaystyle {\frac {\epsilon }{9}}}\\\\\Rightarrow &&&&9|x-2|&<&\epsilon \\\\\Rightarrow &&3|x+2||x-2|&<&9|x-2|&<&\epsilon &&&&&{\textrm {(using~}}\dagger \dagger )\\\\\Rightarrow &&&&|3x^{2}-12|&<&\epsilon \\\\\Rightarrow &&&&|f(x)-L|&<&\epsilon ,\end{array}}}$

as required.

${\displaystyle \square }$

Problem 5. Using the definition of a limit, show that ${\displaystyle {\displaystyle \lim _{x\rightarrow 2}x^{2}+3x+1=11}}$.

Solution. Here, we have ${\displaystyle f(x)=x^{2}+3x+1,\ L=11}$ and ${\displaystyle c=2}$. We again begin with scratchwork, assuming ${\displaystyle |f(x)-L|<\epsilon }$. Then

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

Now, we again assume ${\displaystyle \delta =1}$, so

${\displaystyle -\delta \,=\,-1\,<\,x-2\,<\,1\,=\,\delta ,\qquad \qquad \qquad (\natural )}$

and adding seven to the inequality,

${\displaystyle -\delta \,=\,6\,<\,x+5\,<\,1\,=\,7.}$

Again, dividing ${\displaystyle \epsilon }$ by we have

${\displaystyle {\frac {\epsilon }{7}}\,<\,{\frac {\epsilon }{|x+5|}}\,<\,{\frac {\epsilon }{5}}}$

for all ${\displaystyle x}$ satisfying ${\displaystyle |x-2|<1.}$ We can now write the proof.

Proof. Let ${\displaystyle \epsilon >0}$  be given. Choose ${\displaystyle \delta =\min \left\{1,{\frac {\epsilon }{7}}\right\}}$. Then, whenever ${\displaystyle |x-c|=|x-2|<\delta ,}$

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

as required.

${\displaystyle \square }$

## Other Examples

Problem 6. Using the definition of a limit, show that ${\displaystyle {\displaystyle \lim _{x\rightarrow 1}{\sqrt {x}}=1}}$.

Solution. We have ${\displaystyle f(x)={\sqrt {x}},\ L=1}$  and ${\displaystyle c=1}$. We start our scratchwork as usual, assuming ${\displaystyle |f(x)-L|<\epsilon }$. Then we have

${\displaystyle {\begin{array}{cccrcl}&&&|{\sqrt {x}}-1|&<&{\displaystyle \epsilon }\\\\\Rightarrow &&&|{\sqrt {x}}+1|\cdot |{\sqrt {x}}-1&<&|{\sqrt {x}}+1|\cdot \epsilon \\\\\Rightarrow &&&|x-1|&<&|{\sqrt {x}}+1|\cdot \epsilon .\end{array}}}$

Similar to the quadratic examples, we can set an initial ${\displaystyle \delta =1.}$ Then

${\displaystyle -\delta \,=\,-1\,<\,x-1\,<\,1\,=\,\delta ,}$

and adding one to each term we find

${\displaystyle 0\,<\,x\,<\,2.}$

This, in turn, means that

${\displaystyle 1\,<\,{\sqrt {x}}+1\,<\,{\sqrt {2}}+1.}$

Multiplying the inequality by ${\displaystyle \epsilon }$, we have

${\displaystyle \epsilon \,<\,|{\sqrt {x}}+1|\,<\,\left({\sqrt {2}}+1\right)\epsilon \qquad {\mbox{while}}\qquad {\frac {\epsilon }{{\sqrt {2}}+1}}\,<\,{\frac {\epsilon }{|{\sqrt {x}}+1|}}\,<\,{\frac {\epsilon }{1}}.}$

for all ${\displaystyle x}$ satisfying ${\displaystyle |x-1|<1=\delta }$. This gives us our ${\displaystyle \delta }$, and we can write the proof.

Proof. Let ${\displaystyle \epsilon >0}$ be given. Choose ${\displaystyle \delta =\min \left\{1,\epsilon \right\}}$. Then, whenever ${\displaystyle |x-c|=|x-1|<\delta }$,

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

as required. Notice that we used both of the inequalities involving ${\displaystyle \epsilon }$ and ${\displaystyle \left|{\sqrt {x}}+1\right|}$  to complete the proof.

${\displaystyle \square }$
There are many more difficult examples, but these are meant as an introduction.