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

Formal Definition

We say 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 L} is the limit of 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} at 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} if, for any 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} , there exists a 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>0}   such that 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 0<|x-c|<\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 \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, 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} is really the same thing as

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

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

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

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

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

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

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

From here, we do whatever it takes (usually factoring) to change 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|} into 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|} . Once we have a statement of the form

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|\,<\,\textrm{something},}

this allows us to pick a 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} 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 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}10=10}} .

Solution. Looking at the statement we need to prove, 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)=10,\ L=10}  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} . Since for any 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} , 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)=10} , we know that for any 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} ,

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|\,=\,|10-10|\,=\,0\,<\epsilon}

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

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=1} . 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-1|<\delta=1} , 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)-L|\ =\ |10-10|\ =\ 0\ <\ \epsilon.}

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}

A few quick notes about these types of proofs:

• Every one will begin with "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." 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 </math>\delta</math>. 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 </math>\epsilon</math> and $\left|{\sqrt {x}}+1\right|$ to complete the proof.

</math>\square</math>

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