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

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

In other words, we are trying to restrict $f(x)$ to the $\epsilon$-neighborhood centered at $L$, 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 Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle 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 Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle \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 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

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

</math>\square</math>

A few quick notes about these.

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

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.

</math>\square</math>

It should be mentioned that in cases where Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle f(x)=mx+b} , and $m\neq 0$, we will get that Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle \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 $f(x)=x^{2},\ L=1$ 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 Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle |x-1|} in terms of $\epsilon$, but there's also an $x$ on the right hand side! This requires us to pick an ``initial </math>\delta</math>. Let's choose $\delta =1$. Then, whenever Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle |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 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 Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle \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.

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

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

</math>\square</math>

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 7 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.

</math>\square</math>

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 </math>\epsilon</math> 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.

</math>\square</math>

There are many more difficult examples, but these are meant as an introduction. \end{document}