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

Formal Definition

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

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

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

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

If we then add $L$ to each term, we arrive at

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

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

$c-\delta \,<\,x\,<\,c+\delta .$

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

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

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

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

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

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

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

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

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

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

$|f(x)-L|\ =\ |10-10|\ =\ 0\ <\ \epsilon .$

$\square$

A few quick notes about these types of proofs:

• Every one will begin with "let $\epsilon >0$ be given." That's because most of the time, our $\delta$ will be tied to the $\epsilon$.

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

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

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

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

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

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

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

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

as required.

$\square$

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

Examples with Quadratic Functions

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

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

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

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

$-\delta \,=\,-1\,<\,x-1\,<\,1\,=\,\delta ,$

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

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

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

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

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

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

Now, our proof can be written.

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

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

as required.

</math>\square</math>

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

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

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

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

$-\delta \,=\,-1\,<\,x-2\,<\,1\,=\,\delta .$

Now adding 4 to the inequality, we have

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

Dividing $\epsilon$ by this inequality, we have

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

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

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

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

as required.

</math>\square</math>

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

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

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

Now, we again assume $\delta =1$, so

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

and adding 7 to the inequality,

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

Again, dividing $\epsilon$ by we have

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

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

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

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

as required.

</math>\square</math>

Other Examples

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

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

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

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

$-\delta \,=\,-1\,<\,x-1\,<\,1\,=\,\delta ,$

and adding one to each term we find

$0\,<\,x\,<\,2.$

This, in turn, means that

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

Multiplying the inequality by $\epsilon$, we have

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

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

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

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

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

$\square$

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