The Precise Definition of Limits

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The $ϵ$–$\delta$ Definition of a Limit

Let's revisit the intuitive definition of $\underset{x\to a}{lim}f(x)=L$: 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.

Let's express every element in the above description in mathematical language. 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 $|x-a|$ "sufficiently small," but not zero.

Now, we need to precisely define "as small as we please" and "sufficiently small" mathematically. To achieve this, imagine a game between you and me. You challenge me by giving a small number (as small as you wish) and my task is to find another number such that for all $x\ne a$ satisfying .

Because the inequality is equivalent to

or

the geometrical meaning of this game is as follows: You consider a band of width $2ϵ$ bounded by the lines $y=L-ϵ$ and $y=L+ϵ$ (see the following figure), and I must identify an open interval of radius $\delta$ with a at the center such that all the points on the graph of $y=f(x)$ that correspond to values of x within the interval $(a-\delta ,a+\delta )$—except for the point directly above a—must fall within the band you specified.

Let's explore, with an example, what it means when you give me an $ϵ$, and I can find a corresponding $\delta$.

For example, previously, we saw that if

$$f(x)=\frac{2x^2-2}{x-1},$$

then $\underset{x\to 1}{lim}f(x)=4$. For instance, if you give me $ϵ=0.01$, I will take $\delta =0.005$ (or smaller), and claim for all $x\ne 1$ satisfying ; because if $x\ne 1$ and then

\begin{aligned} |f(x) - 4| &= |2x + 2 - 4| \\ &= |2x - 2| \\ &= 2|x - 1| < 2 \times 0.005 = 0.01. \end{aligned}

Recall that when $x\ne 1$:

$$f(x)=\frac{2(x^2-1)}{x-1}=2\frac{(x-1)(x+1)}{x-1}=2(x+1).$$

If you give me $ϵ=0.0002$, I just need to take $\delta =0.0001$ (or smaller), because and $x\ne 1$ implies that :

\begin{aligned} |f(x) - 4| &= |2x + 2 - 4| \\ &= 2|x - 1| < 2 \times 0.0001 = 0.0002. \end{aligned}

If this game goes on forever and for every you give me, I can find a 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 whenever $|x-a|$ is less than some appropriately chosen positive number $\delta$ and $|x-a|\ne 0$ (because $x\ne a$) then

$$\underset{x\to a}{lim}f(x)=L.$$

Remark that in general the size of $\delta$ depends on the value of $ϵ$.

• Instead of stating " and $|x-a|\ne 0$ (or $x\ne a$)," we can express it as .
• Recall that the condition $x\ne a$ or is imposed because what happens to $f(x)$ when x equals a has no influence on the value or the existence of the limit. The focus is solely on the behavior of $f(x)$ for values of x that are close to a.

Another way of writing the last line is: "for all x:

We use the symbol "$⟹$" in place of "implies" or "if ... then ... ." The above definition is commonly referred to as the epsilon-delta definition of a 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 here.

Also instead of saying "for every , there exists a $\delta$ such that ...", we can say for every neighborhood of $L$, $N_1(L)$, there is some neighborhood of $a$, $N_2(a)$ such that

$$f(x)\in N_1(L)\text{whenever}x\in N_2(a)\text{and}x\ne a.$$

In general:

Example 3. Use the $ϵ$–$\delta$ definition to prove $$\underset{x\to 2}{lim}{x}^2=4.$$

Solution

Here $f(x)=x^2,a=2$, and $L=4$. Suppose is given. We want to find a number such that 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 If we can find a positive constant C such that for x close to 2 and choose $\delta \le ϵ/C$ then 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 or or . If we add 2 to each side and then apply the absolute value we get , so . So if then and Therefore, if we have two restrictions on x, namely then we will have To satisfy both inequalities and , we let $\delta$ be the minimum of 1 and $ϵ/5$: $$\delta \le min\{1,ϵ/5\},$$ and we conclude For example, if $ϵ=0.25$, then we choose $min\{1,0.25/5\}=0.05$ (or less) for $\delta$. The following figure shows when $ϵ=0.25$ and $\delta =0.05$.

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 . In this case ; adding 2 to each side, we get or . Here $4.1\le C$. So if we have two restrictions on x: and , then we will have . Thus another choice for $\delta$ is $\delta \le min\{0.1,ϵ/4.1\}$. In this case, if $ϵ=0.25$ is given, we choose $min\{0.1,0.25/4.1\}\approx 0.061$ (or less) for $\delta$.

Example 5. Use the $ϵ$–$\delta$ definition to prove

Solution

We must show that for every given , there exists a such that for all x: We note that we must have $\delta \le a$, otherwise, $\sqrt{x}$ will not be defined for some x (see the following figure).

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{aligned} |\sqrt{x} - \sqrt{a}| &= \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}{|\sqrt{x} + \sqrt{a}|} |x - a| \end{aligned} Note that because , and (if , $\sqrt{x}$ would be imaginary), . 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 $\left|\frac{A}{B}\right|=\frac{|A|}{|B|}$ ($B\ne 0$). Now we need to find an upper bound for $\frac{1}{|\sqrt{x}+\sqrt{a}|}|x-a|$ and make the upper bound less than $ϵ$. 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 $b\ne 0$, then . That is, If we choose $\delta$ to be less than $ϵ\sqrt{a}$ (and of course less than a), then That is, for all x: if then , where $\delta =min\{a,ϵ\sqrt{a}\}$.

The Precise Definitions of One-Sided Limits

By slightly modifying the definition of two-sided (or ordinary) limits, we obtain the rigorous definitions of one-sided limits.

Similarly

Precise Definition of Infinite Limits

Let's revisit the intuitive meaning of $\underset{x\to a}{lim}f(x)=+\mathrm{\infty }$: If we can make $f(x)$ as large as we wish by taking x sufficiently close to a (but not equal to a), then we say $f(x)$ approaches $+\mathrm{\infty }$ as x approaches a.

Previously, we expressed "taking x sufficiently close to a" in mathematical language using . Now we express "making $f(x)$ as large as we wish" mathematically by saying for any given positive number K.

The precise definition is as follows:

This means $\underset{x\to a}{lim}f(x)=+\mathrm{\infty }$ if for any number that you give me, I can determine a such that for all x closer to a than $\delta$ (except when $x=a$), $f(x)$ lies above the horizontal line $y=K$. A geometric illustration is shown below.

Precise Definitions of Limits at Infinity

Now let's consider limits at infinity, where x becomes arbitrarily large positive ($x\to +\mathrm{\infty }$) or arbitrarily large negative ($x\to -\mathrm{\infty }$).

Recall the intuitive definition of $\underset{x\to +\mathrm{\infty }}{lim}f(x)=L$: The values of $f(x)$ can be made as close to L as we please by taking x sufficiently large.

To express "taking x sufficiently large" in rigorous mathematical language, we say for some positive number N. Similarly, for $x\to -\mathrm{\infty }$, "taking x sufficiently far to the left" means for some negative number N.

\begin{tikzpicture}[>=Stealth, scale=1.1]
  \definecolor{royalblue}{RGB}{54,123,243}
  \definecolor{softblue}{RGB}{220,235,255}

  % Shaded epsilon band
  \fill[softblue] (0, 1.4) rectangle (6.2, 2.6);

  % Axes
  \draw[->, line width=0.8pt] (-0.5,0) -- (6.5,0) node[right] {$x$};
  \draw[->, line width=0.8pt] (0,-0.5) -- (0,3.5) node[above] {$y$};

  % Horizontal lines for L, L+eps, L-eps
  \draw[dashed, royalblue, line width=1pt] (-0.2, 2.0) -- (6.2, 2.0) node[right] {$y=L$};
  \draw[dashed, gray!80, line width=0.8pt] (-0.2, 2.6) -- (6.2, 2.6) node[right] {$y=L+ϵ$};
  \draw[dashed, gray!80, line width=0.8pt] (-0.2, 1.4) -- (6.2, 1.4) node[right] {$y=L-ϵ$};

  % Vertical line for N
  \draw[dashed, red!80!black, line width=1pt] (2.8, 0) -- (2.8, 3.2) node[above] {$x=N$};

  % Ticks and Labels on Y-axis
  \draw[gray] (-0.08, 2.0) -- (0.08, 2.0) node[left=2pt] {$L$};
  \draw[gray] (-0.08, 2.6) -- (0.08, 2.6) node[left=2pt] {$L+ϵ$};
  \draw[gray] (-0.08, 1.4) -- (0.08, 1.4) node[left=2pt] {$L-ϵ$};

  % Tick and Label on X-axis
  \draw[gray] (2.8, -0.08) -- (2.8, 0.08) node[below=2pt] {$N$};

  % Curve y = f(x)
  \draw[royalblue, line width=1.4pt] plot[smooth, domain=0.3:6.0] (\x, {2.0 + 1.25*exp(-0.7*\x)*cos(200*\x r)});
  \node[royalblue, above] at (1.2, 2.7) {$y=f(x)$};

  % Double-headed arrow for epsilon
  \draw[<->, gray!90!black, thick] (5.5, 2.0) -- (5.5, 2.6) node[midway, right] {$ϵ$};
  \draw[<->, gray!90!black, thick] (5.5, 1.4) -- (5.5, 2.0) node[midway, right] {$ϵ$};

  % Region x > N arrow
  \draw[->, line width=1.1pt, red!80!black] (3.0, 0.4) -- (5.2, 0.4) node[midway, above] {};
\end{tikzpicture}

The formal definition is as follows:

Similarly, for limits as $x\to -\mathrm{\infty }$:

Geometrically, $\underset{x\to +\mathrm{\infty }}{lim}f(x)=L$ means that for any horizontal strip of width $2ϵ$ bounded by $y=L-ϵ$ and $y=L+ϵ$, there exists a vertical line $x=N$ such that the graph of $y=f(x)$ stays entirely within the strip for all .

Exercises