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 Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle 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

Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle -\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 Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle {\boldsymbol {\epsilon }}} -neighborhood centered at ${\boldsymbol {L}}$, which is the interval Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle (L-\epsilon ,L+\epsilon )} . Similarly, we can rewrite Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle |x-c|<\delta } to become

Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle 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 Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\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

Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle |f(x)-L|\,<\,\epsilon .}

From here, we do whatever it takes (usually factoring) to change Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle |f(x)-L|} into Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle |x-c|} . Once we have a statement of the form

Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle |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 Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle c=1} . Since for any $x$, we have Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle f(x)=10} , we know that for any $x$,

Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle |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

Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle |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 Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle 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 Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle |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 Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle \delta <\epsilon /2} .

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

Proof. Let $\epsilon >0$  be given. Choose Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle \delta <\epsilon /2} . Then, whenever Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle |x-c|=|x-3|<\delta <\epsilon /2} , we have

Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\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.

$\square$

It should be mentioned that in cases where $f(x)=mx+b$, and Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle 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 Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle {\displaystyle \lim _{x\rightarrow 1}x^{2}=1}} .

Solution. Here, we have Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle f(x)=x^{2},\ L=1}   and Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle c=1} . We again begin with scratchwork. Suppose $|f(x)-L|<\epsilon$. We then solve for Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle |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

Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle 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 Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle |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 Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle b} .

Now, our proof can be written.

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

Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\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.

$\square$

Problem 4. Using the definition of a limit, show that Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle {\displaystyle \lim _{x\rightarrow 2}3x^{2}=12}} .

Solution. This time, Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle f(x)=3x^{2},\ L=12}   and Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle c=2} . We follow the same pattern, doing the scratchwork first. Assume Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle |f(x)-L|<\epsilon} . Then

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\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 Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \delta=1} . Then we have

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle -\delta\,=\,-1\,<\, x-2\,<\,1\,=\,\delta.}

Now adding 4 to the inequality, we have

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 3\,<\, x+2\,<\,5.\qquad\qquad\qquad(\dagger\dagger)}

Dividing Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \epsilon} by this inequality, we have

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{\epsilon}{5}\,<\,\frac{\epsilon}{|x+2|}\,<\,\frac{\epsilon}{3}.}

So we can choose Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\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 Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \epsilon>0}   be given. Choose Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \delta=\min\left\{ 1,\frac{\epsilon}{9}\right\}} . Then, whenever Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle |x-c|=|x-2|<\delta} ,

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\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.

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \square}

Problem 5. Using the definition of a limit, show that Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle {\displaystyle \lim_{x\rightarrow2}x^{2}+3x+1=11}} .

Solution. Here, we have Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f(x)=x^{2}+3x+1,\ L=11} and Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle c=2} . We again begin with scratchwork, assuming Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle |f(x)-L|<\epsilon} . Then

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\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 Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \delta=1} , so

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle -\delta\,=\,-1\,<\, x-2\,<\,1\,=\,\delta,\qquad\qquad\qquad(\natural)}

and adding seven to the inequality,

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle -\delta\,=\,6\,<\, x+5\,<\,1\,=\,7.}

Again, dividing Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \epsilon} by we have

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{\epsilon}{7}\,<\,\frac{\epsilon}{|x+5|}\,<\,\frac{\epsilon}{5}}

for all Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x} satisfying Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle |x-2|<1.} We can now write the proof.

Proof. Let Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \epsilon>0}   be given. Choose Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \delta=\min\left\{ 1,\frac{\epsilon}{7}\right\}} . Then, whenever Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle |x-c|=|x-2|<\delta,}

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\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.

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \square}

Other Examples

Problem 6. Using the definition of a limit, show that Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle {\displaystyle \lim_{x\rightarrow1}\sqrt{x}=1}} .

Solution. We have Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f(x)=\sqrt{x},\ L=1}   and Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle c=1} . We start our scratchwork as usual, assuming Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle |f(x)-L|<\epsilon} . Then we have

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\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 Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \delta=1.} Then

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle -\delta\,=\,-1\,<\, x-1\,<\,1\,=\,\delta,}

and adding one to each term we find

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 0\,<\, x\,<\,2.}

This, in turn, means that

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 1\,<\,\sqrt{x}+1\,<\,\sqrt{2}+1.}

Multiplying the inequality by Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \epsilon} , we have

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\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 Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x} satisfying Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle |x-1|<1=\delta} . This gives us our Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \delta} , and we can write the proof.

Proof. Let Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \epsilon>0} be given. Choose Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \delta=\min\left\{ 1,\epsilon\right\}} . Then, whenever Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle |x-c|=|x-1|<\delta} ,

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\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 Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \epsilon} and Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \left|\sqrt{x}+1\right|}   to complete the proof.

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \square}

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