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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.
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The 𝝐–𝞭 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 $a$.
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.
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- 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.
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| 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.
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 $\delta=0.05$.
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| 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 $\delta$.
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).
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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 $\delta=\min\{a,\epsilon\sqrt{a}\}$.