Definition of a Limit

The ideas of the previous examples and practice problems can be stated for general functions and limits, and they provide the basis for the definition of limit which is given in the box. The use of the lower case Greek letters $\varepsilon$ (epsilon) and $\delta$ (delta) in the definition is standard, and this definition is sometimes called the "epsilon-delta" definition of limit.

In this definition, $\varepsilon$ represents the given distance on either side of the limiting value $\mathrm{y}=\mathrm{L}$, and $\delta$ is the distance on each side of the point $x=a$ on the $x$-axis that we have been finding in the previous examples. This definition has the form of $\mathrm{a}$ a "challenge and reponse:" for any positive challenge $\varepsilon$ (make $f(c)$ within $\varepsilon$ of $\mathrm{L}$), there is a positive response $\delta$ (start with $\mathrm{x}$ within $\delta$ of $\mathrm{a}$ and $x \neq a$).

Example 4: In Fig. 11a, $\lim\limits_{\mathrm{x} \rightarrow \mathrm{a}} \mathrm{f}(\mathrm{x})=\mathrm{L}$, and a value for $\varepsilon$ is given graphically as a length. Find a length for $\delta$ that satisfies the definition of limit (so "if $\mathrm{x}$ is within $\delta$ of $\mathrm{a}($ and $\mathrm{x} \neq \mathrm{a})$, then $\mathrm{f}(\mathrm{x})$ is within $\varepsilon$ of $\mathrm{L}$").

Solution: Follow the steps outlined in Example 3. The length for $\delta$ is shown in Fig. 11b, and any shorter length for $\delta$ also satisfies the definition.

Practice 4: In Fig. 12, $\lim\limits_{\mathrm{x} \rightarrow \mathrm{a}} \mathrm{f}(\mathrm{x})=\mathrm{L}$, and a value for $\varepsilon$ is given graphically as a length. Find a length for $\delta$ that satisfies the definition of limit.

Example 5: Prove that $\lim\limits_{\mathrm{x} \rightarrow 3} 4 \mathrm{x}-5=7$

Solution: We need to show that

"for every given $\varepsilon > 0$ there is a $\delta > 0$ so that

Actually there are two things we need to do. First, we need to find a value for $\delta$ (typically depending on $\varepsilon$), and, second, we need to show that our $\delta$ really does satisfy the "if $-$ then" part of the definition.

i. Finding $\delta$ is similar to part (c) in Example 1 and Practice 1: assume $4 \mathrm{x}-5$ is within $\varepsilon$ units of 7 and solve for $x$. If $7-\varepsilon < 4 x-5 < 7+\varepsilon$, then $12-\varepsilon < 4 x < 12+\varepsilon$ and $3-\varepsilon / 4 < x < 3+\varepsilon / 4$, so $x$ is within $\varepsilon / 4$ units of 3. Put $\delta=\varepsilon / 4$

ii. To show that $\delta=\varepsilon / 4$ satisfies the definition, we merely reverse the order of the steps in part i. Assume that $x$ is within $\delta$ units of 3. Then $3-\delta < x < 3+\delta$ so

$\begin{array}{ll}3-\varepsilon / 4 < \mathrm{x} < 3+\varepsilon / 4 & (\text { replacing } \delta \text { with } \varepsilon / 4) \\ 12-\varepsilon < 4 \mathrm{x} < 12+\varepsilon & \text { (multiplying by 4), and } \\ 7-\varepsilon < 4 \mathrm{x}-5 < 7+\varepsilon & \text { (subtracting 5), so }\end{array}$ 

we can conclude that $\mathrm{f}(\mathrm{x})=4 \mathrm{x}-5$ is within $\varepsilon$ units of 7. This formally verifies that $\lim\limits_{\mathrm{x} \rightarrow 3} 4 \mathrm{x}-5=7$.

Practice 5: Prove that $\lim\limits_{x \rightarrow 4} 5 x+3=23$.

The method used to prove the values of the limits for these particular linear functions can also be used to prove the following general result about the limits of linear functions.

Theorem: $\quad \lim\limits_{\mathrm{x} \rightarrow \mathrm{a}} \mathrm{m} \mathrm{x}+\mathrm{b}=\mathrm{ma}+\mathrm{b}$

Proof: Let $f(x)=m x+b$.

Case 1: $\mathrm{m}=0$. Then $\mathrm{f}(\mathrm{x})=0 \mathrm{x}+\mathrm{b}=\mathrm{b}$ is simply a constant function, and any value for $\delta > 0$ satisfies the definition. Given any value of $\varepsilon > 0$, let $\delta=1$ (any positive value for $\delta$ works). If $x$ is is within 1 unit of $\mathrm{a}$, then $f(x)-f(a)=b-b=0 < e$, so we have shown that for any $\varepsilon > 0$, there is a $\delta > 0$ which satisfies the definition.

Case 2: $\mathrm{m} \neq 0$. Then $f(x)=m x+b$. For any $\varepsilon > 0$, put $\delta=\frac{\varepsilon}{\mid m \mid} > 0$. If $x$ is within $\delta=\frac{\varepsilon}{\mid m \mid}$ of $\mathrm{a}$, then

$\mathrm{a}-\frac{\varepsilon}{\mid m \mid} < \mathrm{x} < \mathrm{a}+\frac{\varepsilon}{|\mathrm{m}|} \quad \text { so}-\frac{\varepsilon}{\mid m \mid} < \mathrm{x}-\mathrm{a} < \frac{\varepsilon}{\mid m \mid} \quad$  and $|\mathrm{x}-\mathrm{a}| < \frac{\varepsilon}{\mid m \mid}$

Then the distance between $\mathrm{f}(\mathrm{x})$ and $\mathrm{L}=\mathrm{ma}+\mathrm{b}$ is $|\mathrm{f}(\mathrm{x})-\mathrm{L}|=|(\mathrm{m} \mathrm{x}+\mathrm{b})-(\mathrm{ma}+\mathrm{b})|=\mid m|\cdot| \mathrm{x}-\mathrm{a}| < \mid m| \frac{\varepsilon}{\frac{\varepsilon}{\mathrm{m} \mid}}=\varepsilon$ so $\mathrm{f}(\mathrm{x})$ is within $\varepsilon$ of $\mathrm{L}=\mathrm{ma}+\mathrm{b}$. (Fig. 13)

In each case, we have shown that "given any $\varepsilon > 0$, there is a $\delta > 0$" that satisfies the rest of the definition is satisfied.

If there is even a single value of $\varepsilon$ for which there is no $\delta$, then the function does not satisfy the definition, and we say that the limit "does not exist".

Example 6: Let $f(x)=\left\{\begin{array}{ll}2 & \text { if } x < 1 \\ 4 & \text { if } x > 1\end{array}\right.$ as is shown in Fig. 14.

Use the definition to prove that $\lim\limits_{x \rightarrow 1} f(x)$ does not exist.

Solution: One common proof technique in mathematics is called "proof by contradiction," and that is the method we use here. Using that method in this case, (i) we assume that the limit does exist and equals some number $\mathrm{L}$, (ii) we show that this assumption leads to a contradiction, and (iii) we conclude that the assumption must have been false. Therefore, we conclude that the limit does not exist.

(i) Assume that the limit exists: $\quad \lim\limits_{\mathrm{x} \rightarrow 1} \mathrm{f}(\mathrm{x})=\mathrm{L}$ for some value for $\mathrm{L}$. Let $\varepsilon=\frac{1}{2}$. (The definition says "for every ε" so we can pick this value. Why we chose this value for $\varepsilon$ shows up later in the proof). Then, since we are assuming that the limit exists, there is a $\delta > 0$ so that if $\mathrm{x}$ is within $\delta$ of 1 then $\mathrm{f}(\mathrm{x})$ is within $\varepsilon$ of $\mathrm{L}$.

(ii) Let $x_{1}$ be between 1 and $1+\delta$. Then $x_{1} > 1$ so $f\left(x_{1}\right)=4$. Also, $x_{1}$ is within $\delta$ of 1 so $f\left(x_{1}\right)=$ 4 is within $\frac{1}{2}$ of $\mathrm{L}$, and $\mathrm{L}$ is between $3.5$ and 4.5: $3.5 < \mathbf{L} < 4.5$.

Let $x_{2}$ be between 1 and $1-\delta$. Then $x_{2} < 1$ so $f\left(x_{2}\right)=2$. Also, $x_{2}$ is within $\delta$ of 1 so $f\left(x_{2}\right)=$ 2 is within $\frac{1}{2}$ of $\mathrm{L}$, and $\mathrm{L}$ is between $1.5$ and $2.5: \mathbf{1. 5} < \mathbf{L} < \mathbf{2. 5}$.

(iii) The two inequalities in bold print provide the contradiction we were hoping to find. There is no value $\mathrm{L}$ that simultaneously satisfies $3.5 < \mathrm{L} < 4.5$ and $1.5 < \mathrm{L} < 2.5$, so we can conclude that our assumption was false and that $f(x)$ does not have a limit as $x \rightarrow 1$.

Practice 6: Use the definition to prove that $\lim\limits_{x \rightarrow 0} \frac{1}{x}$ does not exist (Fig. 15).

Two Limit Theorems

The theorems and their proofs are included here so you can see how such proofs proceed - you have already used these theorems to evaluate limits of functions. There are rigorous proofs of all of the other limit properties, but they are somewhat more complicated than the proofs given here.

Theorem: If $\lim\limits_{\mathrm{x} \rightarrow \mathrm{a}} \mathrm{f}(\mathrm{x})=\mathrm{L}$, then $\lim\limits_{\mathrm{x} \rightarrow \mathrm{a}} \mathrm{k} \cdot \mathrm{f}(\mathrm{x})=\mathrm{k} \cdot \mathrm{L}$.

Proof: Case $\mathrm{k}=0$: The Theorem is true but not very interesting: $\quad \lim\limits_{\mathrm{x} \rightarrow \mathrm{a}} 0 \cdot \mathrm{f}(\mathrm{x})=\lim\limits_{\mathrm{x} \rightarrow \mathrm{a}} 0=0 \cdot \mathrm{L}$.

Case $k \neq 0$: Since $\lim\limits_{x \rightarrow a} f(x)=L$, then, by the definition, for every $\varepsilon > 0$ there is a $\delta > 0$ so that $|f(x)-L| < \varepsilon$ whenever $|x-a| < \delta.$ For any $\varepsilon > 0$, we know $\frac{\varepsilon}{|k|} > 0$ and pick a value of $\delta$ that satisfies $|f(x)-L| < \frac{\varepsilon}{|k \mid}$ whenever $|x-a| < \delta$. When 

$|x-a| < \delta \quad("x$ is within $\delta$ of $\mathrm{a}$") then $|f(x)-L| < \frac{\varepsilon}{|k|} \quad\left("f(x)\right.$ is within $\frac{\varepsilon}{|k|}$ of L") so $|k| \cdot|f(x)-L| < \varepsilon \quad$ (multiplying each side by $|\mathrm{k}| > 0$) and $|\mathrm{k} \cdot \mathrm{f}(\mathrm{x})-\mathrm{k} \cdot \mathrm{L}| < \varepsilon \quad(\mathrm{k} \cdot \mathrm{f}(\mathrm{x})$ is within $\varepsilon$ of $\mathrm{k} \cdot \mathrm{L})$

Theorem: If $\lim\limits_{x \rightarrow a} f(x)=L$ and $\lim\limits_{x \rightarrow a} g(x)=M$, then $\lim\limits_{x \rightarrow a} f(x)+g(x)=L+M$.

Proof: Assume that $\lim\limits_{x \rightarrow a} f(x)=L$ and $\lim\limits_{x \rightarrow a} g(x)=M$. Then, given any $\varepsilon > 0$, we know $\varepsilon / 2 > 0$ and that there are deltas for $\mathrm{f}$ and $\mathrm{g}, \delta_{\mathrm{f}}$ and $\delta_{\mathrm{g}}$, so that 

if $|x-a| < \delta_{f}$, then $|f(x)-L| < \varepsilon / 2$ ("if $x$ is within $\delta_{f}$ of $a$, then $f(x)$ is within $\varepsilon / 2$ of $L$", and 

if $|x-a| < \delta_{g}$, then $\lg (x)-M \mid < \varepsilon / 2$ ("if $x$ is within $\delta_{g}$ of $\mathrm{a}$, then $g(x)$ is within $\varepsilon / 2$ of $\mathrm{M}$").

Let $\delta$ be the smaller of $\delta_{f}$ and $\delta_{g}.$ If $|x-a| < \delta$, then $|f(x)-L| < \varepsilon / 2$ and $|g(x)-M| < \varepsilon / 2$ so

$\mid(f(x)+g(x))-(L+M))|=|(f(x)-L)+(g(x)-M) \mid \quad$ (rearranging the terms)

$\leq |f(x)-L|+|g(x)-M| \quad$ (by the Triangle Inequality for absolute values)

$< \frac{\varepsilon}{2}+\frac{\varepsilon}{2}=\varepsilon. \quad$ (by the definition of the limits for $\mathrm{f}$ and $\mathrm{g}$).