4.2 The Precise Definition of a Limit

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So far, we have defined the limit of a function $f$ as $x$ approaches $a$ qualitatively. In this section, we are going to introduce the definition of limit on a sound mathematical basis.

History

 

Read about the definition history of a limit

Although mathematicians intuitively applied limiting processes even before the development of calculus, without a precise definition of limit, they were not able to prove important theorems of calculus with sufficient rigor. The first person who tried to put the definition on a mathematical sound basis was the French mathematician, engineer, and physicist, Augustin-Louis Cauchy (1789–1857). Finally, the definitive modern definition of limit was formulated by the German mathematician Karl Weierstrass (1815–1897) who used two greek letters $\epsilon$ (epsilon) and $\delta$ (delta) for the small differences. For more information read this Wikipedia article.

 

The ????–???? Definition of Limit

Let’s review the informal definition of the limit of a function that says the limit of $f(x)$ as $x$ approaches $a$ is $L$ if we can make the values of $f(x)$ as close to $L$ as we please by taking $x$ sufficiently close to $a$, but not equal to $a$.

We say two quantities A and B are close when the distance between them is small. Because the distance from $x$ to $a$ is $|x-a|$ and the distance from $f(x)$ to $L$ is $|f(x)-L|$, we can alternatively say that the limit of $f(x)$ as $x$ approaches $a$ is $L$ if we can make $|f(x)-L|$ “as small as we please” by making $|x-a|$ “small enough”, but not zero.
 
Now we need to make “as small as we please” and “small enough” mathematically exact. To this end, consider a game between you and me. You challenge me by giving a small number $\epsilon>0$ (as small as you please) and I have to find another number $\delta>0$ such that $|f(x)-L|<\epsilon$ for all $x\neq a$ satisfying $|x-a|<\delta$.
 

Because the inequality $|f(x)-L|<\epsilon$ is equivalent to

\[-\epsilon<f(x)-L<\epsilon\]

or

\[L-\epsilon<f(x)<L+\epsilon,\]

the geometrical meaning of this game is: you consider a band of width $2\epsilon$ bounded by the lines $y=L-\epsilon$ and $y=L+\epsilon$ (see Figure 1), and I need to find an open interval of radius $\delta$ with $a$ at the center such that the entire points on the graph of $y=f(x)$ above the interval $(-a-\delta,a+\delta)$—except possibly the point above $a$ itself— lie within the band you gave me.

    • Recall that an open interval with $a$ at the center is called a neighborhood of $a$. For more information see the Section on Absolute Value.
Figure 1

For example, at the beginning of the Section on the Concept of a Limit, we saw that ${\displaystyle \lim_{x\to1}f(x)=4}$ where

\[f(x)=\frac{4x^{2}-4}{2x-2}.\]

For instance, if you give me $\epsilon=0.01$, I will take $\delta=0.005$ (or smaller), and claim $|f(x)-4|<0.01$ for all $x\neq1$ satisfying $|x-1|<0.005$; because if $x\neq1$ and $|x-1|<0.005$ then

\begin{align*}
|f(x)-4| & =|2x+2-4|\\
& =|2x-2|\\
& =2|x-1|<2\times0.005=0.01.
\end{align*}

Recall that when $x\neq1$\[f(x)=\frac{4(x^{2}-1)}{2(x-1)}=2\frac{(x-1)(x+1)}{(x-1)}=2(x+1).\]

 

If you give me $\epsilon=0.0002$, I just need to take $\delta=0.0001$ (or smaller), because $|x-1|<0.0001$ and $x\neq1$ implies that $|f(x)-4|<0.0002$

\begin{align*}
|f(x)-4| & =|2x+2-4|\\
& =2|x-1|<2\times0.0001=0.0002.
\end{align*}

If this game goes on for ever and for every $\epsilon>0$ you give me, I can find a $\delta>0$ with the aforementioned conditions, then we say the limit of $f(x)$ as $x$ approaches $a$ is $L$. Specifically, we state that if we can make $|f(x)-L|$ less than any given positive number $\epsilon>0$ whenever $|x-a|$ is less than some appropriately chosen positive number $\delta$ and $|x-a|\neq0$ (because $x\neq a$) then

\[\lim_{x\to a}f(x)=L.\]

We remark that in general the size of $\delta$ depends on the size of $\epsilon$.

Instead of writing $|x-a|<\delta$ and $|x-a|\neq0$ (or $x\neq a$), we can concisely write \[0<|x-a|<\delta.\]

Recall that the condition $x\neq a$ or $0<|x-a|$ is imposed because we deal with the values of $f(x)$ for $x$ close to $a$ not equal to $a$, and the exact value of $f(x)$ at $x=a$ has no influence on the value or the existence of the limit.

Definition 1: Let $f$ be a function that is defined at every number in some open interval containing $a$ except possibly at the number $a$ itself. The symbol ${\displaystyle \lim_{x\to a}f(x)=L}$ means that for every $\epsilon>0$, however small, there exists a $\delta>0$ such that \[|f(x)-L|<\epsilon\qquad\text{whenever}\qquad0<|x-a|<\delta.\]

Another way of writing the last line is: 

\[ \bbox[#F2F2F2,5px,border:2px solid black]{\large \text{“for all } x: \quad 0<|x-a|<\delta\implies|f(x)-L|<\epsilon\text{ ”}}\]

We use the symbol  “⟹” in place of “implies” or “if … then … .”

The above definition is often called the epsilon-delta definition of  limit.

Instead of saying “let $f$ be a function that is defined at every number in some open interval containing $a$ except possibly at the number $a$ itself”, we can say “let $f$ be defined in a deleted neighborhood of the point $a$.” For the definition of the deleted neighborhood, see the Section on Absolute Value.

Example 1

Let the function $f$ be defined by $f(x)=5x+1$. Given ${\displaystyle \lim_{x\to0}f(x)=1}$, find a $\delta$ such that

\[|f(x)-1|<0.01\qquad\text{whenever}\qquad 0<|x|<\delta.\]

Solution

Here $\epsilon=0.01$ is given. To find a $\delta$, we need to establish a connection between

\[|f(x)-L|=|(5x+1)-1|\quad\text{and}\quad|x-a|=|x|.\]

We notice that

\[|f(x)-1|=|5x+1-1|=5|x|\]

Therefore, we want

\[5|x|<0.01\qquad\text{whenever}\qquad0<|x|<\delta\]

If we take $\delta=0.01/5=0.002$, then we have

\[5|x|<5\times0.002=0.01\qquad\text{whenever} \qquad 0<|x|<0.002\]

Note that $0.002$ is the largest value that we can choose for $\delta$; any number less than 0.002 for $\delta$ also works. That is, if $0<\alpha<0.002$, then

\[|f(x)-1|<0.01,\qquad\text{whenever}\qquad 0<|x-0|<\alpha.\]

because any number $x$ that satisfies $0<|x-0|<\alpha$ also satisfies $0<|x-0|<0.002.$

To prove using the $\epsilon-\delta$ definition, we suppose $\epsilon>0$ is given. To find an appropriately chosen $\delta$, we need to establish a connection between $0<|x-a|<\delta$ and $|f(x)-L|<\epsilon$. To this end, work backward from the inequality $|f(x)-L|<\epsilon$, we simplify it that until we get $|x-a|<g(\epsilon)$, where $g(\epsilon)$ is a function of $\epsilon$, then we let $\delta\le g(\epsilon)$.

Example 2

Use the $\epsilon-\delta$ definition to prove
\[\lim_{x\to1}(2x-3)=-1.\]

Solution

Here$f(x)=2x-3$, $a=1$, and $L=-1$. Suppose $\epsilon>0$ is given. We want to find a number $\delta>0$ such that

\[0<|x-1|<\delta\Rightarrow|2x-3-(-1)|<\epsilon.\]

We start with $|2x-3+1|<\epsilon$ and simplify it

\[|2x-3+1|=|2x-2|=2|x-1|<\epsilon\]

or equivalently

\[|x-1|<\frac{1}{2}\epsilon.\]

So if we choose $\delta\le\epsilon/2$, then

\[0<|x-1|<\delta\le\epsilon/2\Rightarrow|2x-3-(-1)|<\epsilon.\]

Example 3

Use the $\epsilon-\delta$ definition to prove

\[\lim_{x\to2}x^{2}=4.\]

Solution

Here $f(x)=x^{2},a=2$, and $L=4$. Suppose $\epsilon>0$ is given. We want to find a number $\delta>0$ such that

\[0<|x-2|<\delta\Rightarrow|x^{2}-4|<\epsilon\]

To establish a connection between $|x^{2}-4|$ and $|x-2|$, we factor $x^{2}-4=(x-2)(x+2).$

\[|x^{2}-4|=|(x-2)(x+2)|=|x-2|\ |x+2|.\]

So we want

\[0<|x-2|<\delta\implies|x-2|\ |x+2|<\epsilon.\]

If we can find a positive constant $C$ such that

\[|x+2|<C\]

for $x$ close to $2$ and choose $\delta\le\epsilon/C$ then

\[0<|x-2|<\frac{\epsilon}{C}\implies|x-2|\ |x+2|<\frac{\epsilon}{C}\ C=\epsilon.\]

So the question is: how can we find $C$? We can find such a number $C$ if we restrict $x$ to some neighborhood of $2$ (= an interval with center at $2$). For example, because $x$ gets closer and closer to 2, we can assume that $x$ is restricted to the numbers that are closer to 2 than 1 unit; that is $|x-2|<1$ or

\[-1<x-2<1\]

or

\[1<x<3.\]

If we add $2$ to each side and then apply the absolute value we get

\[3<x+2<5\]

\[3<|x+2|<5.\]

So if $|x-2|<1$ then $|x+2|<5$ and

\[|x^{2}-4|=|x-2|\ |x+2|<5|x-2|\]

Therefore, if we have two restrictions on $x$, namely

\[|x-2|<1\qquad\text{and}\qquad|x-2|<\frac{1}{5}\epsilon,\tag{i}\]

then we will have

\[|x^{2}-4|<5|x-2|<\epsilon.\]

To satisfy both inequalities $|x-2|<1$ and $|x-2|<\epsilon/5$, we let $\delta$ be the minimum of 1 and $\epsilon/5$

\[\delta\leq\min\left\{ 1,\epsilon/5\right\} ,\]

and we conclude

\[0<|x-2|<\delta\implies|x^{2}-4|<\epsilon.\]

For example, if $\epsilon=0.25$, then we choose $\min\{1,0.25/5\}=0.05$ (or less) for $\delta$. The following figure shows when $\epsilon=0.25$ and $\delta=0.05$.

Figure 2

Notice that somebody else might say because $x$ has to be sufficiently close to 2, we should concern ourselves to only those values of $x$ that are not farther away from 2 than 0.1 units; that is, the values of $x$ for which $|x-2|<0.1$. In this case

\[1.9<x<2.1;\]

adding 2 to each side, we get

\[3.9<x+2<4.1\]

or

\[3.9<|x+2|<4.1.\]

Here $4.1\leq C$. So if we have two restrictions on $x$

\[|x-2|<0.1\quad\text{and}\quad|x-2|<\epsilon/4.1\]

then we will have

\[|x^{2}-4|<\epsilon.\]

Thus another choice for $\delta$ is $\delta\leq\min\left\{ 0.1,\epsilon/4.1\right\} $. In this case, if $\epsilon=0.25$ is given, we choose $\min\{0.1,0.25/4.1\}\approx0.061$ (or less) for $\delta$.

Example 4

Use the $\epsilon-\delta$ definition to prove

\[\lim_{x\to\frac{1}{2}}\frac{1}{x^{2}}=4.\]

Solution

Here $f(x)=1/x^{2},a=1/2$, and $L=4$. We need to show that for every $\epsilon>0$, there exists $\delta>0$ such that 

\[0<\left|x-\frac{1}{2}\right|<\delta\implies\left|\frac{1}{x^{2}}-4\right|<\epsilon.\]

Suppose $\epsilon>0$ is given. To establish a connection between $|x-1/2|$ and $|1/x^{2}-4|$, we work backward from $|1/x^{2}-4|$ and simplify it:

\begin{align*}
\left|\frac{1}{x^{2}}-4\right| & =\left|\frac{1-4x^{2}}{x^{2}}\right|\\
& =\left|\frac{4(\frac{1}{4}-x^{2})}{x^{2}}\right|\\
& =\frac{4}{x^{2}}\left|\frac{1}{2}-x\right|\left|\frac{1}{2}+x\right|\\
& =\frac{4}{x^{2}}\left|x-\frac{1}{2}\right|\left|x+\frac{1}{2}\right|
\end{align*}

Here we have factored $\frac{1}{4}-x^{2}=\left(\frac{1}{2}\right)^{2}-x^{2}$ using the Difference of Square formula (see the Section on Factorization). Now we need to find an upper bound for $4|x+1/2|/x^{2}$ when $x$ is close to $1/2$; that is, to find a constant $C>0$ such that for $x$ close to $1/2$

\[\frac{4}{x^{2}}\left|x+\frac{1}{2}\right|<C.\]

Similar to the previous example, we proceed by restricting $x$ to lie in some neighborhood of $1/2$ (= an interval centered at $1/2$), but unlike the previous example, the radius of the neighborhood cannot be 1 because $\frac{4}{x^{2}}\left|x+\frac{1}{2}\right|$ is not defined for $x=0$. So we consider a neighborhood such that it does not include $x=0$. For example, we assume that $x$ is within a distance $1/4$ from $1/2$; that is,

\[\left|x-\frac{1}{2}\right|<\frac{1}{4},\]

or

\[\frac{1}{4}<x<\frac{3}{4}.\]

When $1/4<x<3/4$

\[\frac{3}{4}<x+\frac{1}{2}<\frac{5}{4}\]

and

\[\frac{1}{16}<x^{2}<\frac{9}{16}.\]

Therefore in this neighborhood of $1/2$:

\[\left|x+\frac{1}{2}\right|<\frac{5}{4},\]

\[\frac{16}{9}<\frac{1}{x^{2}}<16\]

and finally

\[\frac{4}{x^{2}}\left|x+\frac{1}{2}\right|<4\times16\times\frac{5}{4}=80.\]

This means that if $|x-\frac{1}{2}|<\frac{1}{4}$, then

\[\left|\frac{1}{x^{2}}-4\right|=\frac{4}{x^{2}}\left|x-\frac{1}{2}\right|\left|x+\frac{1}{2}\right|<80\left|x-\frac{1}{2}\right|.\]

If $|x-\frac{1}{2}|<\epsilon/80$, then

\[80\left|x-\frac{1}{2}\right|<\epsilon.\]

If we take $\delta\leq\min\left\{ \frac{1}{4},\frac{\epsilon}{80}\right\} $, then whenever $|x-\frac{1}{2}|<\delta$, certainly $|x-\frac{1}{2}|<\frac{1}{4}$ and

\[\left|\frac{1}{x^{2}}-4\right|<80\left|x-\frac{1}{2}\right|.\]

Additionally because $|x-\frac{1}{2}|<\frac{\epsilon}{80}$, we will have

\[\left|\frac{1}{x^{2}}-4\right|<80\left|x-\frac{1}{2}\right|<\epsilon.\]

So we showed that for every $\epsilon>0$, we can take $\delta\leq\min\left\{ \frac{1}{4},\frac{\epsilon}{80}\right\} $ and

\[\left|\frac{1}{x^{2}}-4\right|<\epsilon\quad\text{whenever}\quad\left|x-\frac{1}{2}\right|<\delta.\]

Example 5

Use the $\epsilon$-$\delta$ definition to prove

\[\lim_{x\to a}\sqrt{x}=\sqrt{a},\qquad(a>0)\]

Solution

We must show that for every given $\epsilon>0$, there exists a $\delta>0$ such that for all $x$

\[ \text{if }|x-a|<\delta\text{ then }|\sqrt{x}-\sqrt{a}|<\epsilon \]

We note that we must have $\delta\leq a$, otherwise, $\sqrt{x}$ will not be defined for some $x$ (see Figure 3).

Figure 3: Graph of $y=\sqrt{x}$. If $\delta>a$, then as you can see in the above figure for some $x$ between $a-\delta$ and $0$ (in the above figure such $x$-value is denoted by $x^*$), $\sqrt{x}$ is not defined. Therefore, we must always choose $\delta\leq a$.

To express $|\sqrt{x}-\sqrt{a}|$ in terms of $|x-a|$, we multiply and divide $\sqrt{x}-\sqrt{a}$ by its conjugate $\sqrt{x}+\sqrt{a}$.

\begin{align*}
\left|\sqrt{x}-\sqrt{a}\right| & =\left|(\sqrt{x}-\sqrt{a})\frac{\sqrt{x}+\sqrt{a}}{\sqrt{x}+\sqrt{a}}\right|\\
& =\left|\frac{x-a}{\sqrt{x}+\sqrt{a}}\right|\\
& =\frac{1}{\left|\sqrt{x}+\sqrt{a}\right|}|x-a|
\end{align*}

Note that because $a>0$, and $x>0$ (if $x<0$, $\sqrt{x}$ would be imaginary), $\sqrt{x}+\sqrt{a}>0$. For the second line of the above equation, we used the identity $(A-B)(A+B)=A^{2}-B^{2}$, and the last line follows from the fact that

\[\left|\frac{A}{B}\right|=\frac{|A|}{|B|}\qquad(B\neq0)\]

(See property 6 in the Section on Absolute Value). 

Now we need to find an upper bound for

\[\frac{1}{\left|\sqrt{x}+\sqrt{a}\right|}|x-a|\]

and make the upper bound less than $\epsilon$. We get an upper bound for this fraction if $x$ is replaced by $0$ in the denominator, because if $x$ is replaced by any other number $b\ne0$, then $0+\sqrt{a}<\sqrt{b}+\sqrt{a}$. That is,

\[\frac{1}{\left|\sqrt{x}+\sqrt{a}\right|}|x-a|<\frac{1}{\sqrt{a}}|x-a|.\]

If we choose the $\delta$ to be less than $\epsilon\sqrt{a}$ (and of course less than $a$), then

\[|x-a|<\delta<\epsilon\sqrt{a}\Rightarrow|\sqrt{x}-\sqrt{a}|=\frac{1}{\left|\sqrt{x}+\sqrt{a}\right|}|x-a|<\frac{1}{\sqrt{a}}|x-a|<\frac{1}{\sqrt{a}}\epsilon\sqrt{a}=\epsilon\]

That is, for all $x$

\[\text{if }|x-a|<\delta\text{ then }|\sqrt{x}-\sqrt{a}|<\epsilon,\]

where $\delta=\min\{a,\epsilon\sqrt{a}\}$.