Definition of a Limit

The precise ("formal") definition of limit carefully defines the ideas that we have already been using graphically and intuitively. The following side–by–side columns show some of the phrases we have been using to describe limits, and those phrases, particularly the last ones, provide the basis to building the definition of limit. 

Let's examine what the last phrase ("we can.."). means for the Particular Limit.

Example 1: We know $\lim\limits_{\mathrm{x} \rightarrow 3} 2 \mathrm{x}-1=5$. Show that we can guarantee that the values of $\mathrm{f}(\mathrm{x})=2 \mathrm{x}-1$ are as close to 5 as we want by starting with values of $x$ sufficiently close to 3.

(a) What values of $x$ guarantee that $f(x)=2 x-1$ is within $\mathrm{1}$ unit of $\mathrm{5}$? (Fig. 1a)

Solution: "within $\mathrm{1}$ unit of $\mathrm{5}$" means between $5-1=4$ and $5+1=6$, so the question can be rephrased as "for what values of $x$ is $y=2 x-1$ between 4 and $6: 4 < 2 x-1 < 6$? We want to know which values of $x$ put the values of $y=2 x-1$ into the shaded band in Fig. 1a. The algebraic process is straightforward: solve $4 < 2 x-1 < 6$ for $x$ to get $5 < 2 x < 7$ and $2.5 < x < 3.5$. We can restate this result as follows: "If $\mathrm{x}$ is within $\mathbf{0. 5}$ units of 3, then $\mathrm{y}=2 \mathrm{x}-1$ is within $\mathrm{1}$ unit of $\mathrm{5}$". (Fig. 1b) 

Any smaller distance also satisfies the guarantee: e.g., "If $x$ is within $0.4$ units of 3, then $y=2 x-1$ is within 1 unit of 5". (Fig. 1c)

(b) What values of $x$ guarantee the $f(x)=2 x-1$ is within $0.2$ units of $5$? (Fig. 2a)

Solution: "within $0.2$ units of 5" means between $5-0.2=4.8$ and $5+0.2=5.2$, so the question can be rephrased as "for what values of $x$ is $y=2 x-1$ between $4.8$ and 5.2: $4.8 < 2 x-1 < 5.2$?" Solving for $x$, we get $5.8 < 2 x$ $< 6.2$ and $2.9 < x < 3.1$. "If $x$ is within $0.1$ units of 3, then $y=2 x-1$ is within $0.2$ units of $5.$" (Fig. 2b) Any smaller distance also satisfies the guarantee.

Rather than redoing these calculations for every possible distance from 5, we can do the work once, generally:

(c) What values of $x$ guarantee that $f(x)=2 x-1$ is within $E$ units of 5? (Fig. 3a)

Solution: "within $\mathrm{E}$ unit of 5" means between $5-\mathrm{E}$ and $5+\mathrm{E}$, so the question is "for what values of $x$ is $y=2 x-1$ between $5-E$ and $5+\varepsilon: 5-\mathrm{E} < 2 \mathrm{x}-1 < 5+\mathrm{E}$?" Solving $5-\mathrm{E} < 2 \mathrm{x}-1 < 5+\mathrm{E}$ for $\mathrm{x}, \mathrm{we}$ get $6-\mathrm{E} < 2 \mathrm{x} < 6+\mathrm{E}$ and $3-\mathrm{E} / 2 < \mathrm{x} < 3+\mathrm{E} / 2$. "If $\mathrm{x}$ is within $\mathrm{E} / 2$ units of 3, then $y=2 x-1$ is within $E$ units of 5". (Fig. 3b) Any smaller distance also satisfies the guarantee.

Part (c) of Example 1 illustrates a little of the power of general solutions in mathematics. Rather than doing a new set of similar calculations every time someone demands that $\mathrm{f}(\mathrm{x})=2 \mathrm{x}-1$ be within some given distance of 5, we did the calculations once. And then we can respond for any given distance. For the question "What values of $x$ guarantee that $f(x)=2 x-1$ is within $0.4,0.1$ and $0.006$ units of 5?", we can answer "If $\mathrm{x}$ is within $0.2(=0.4 / 2), 0.05(=0.1 / 2)$ and $0.003(=0.006 / 2)$ units of 3".

Practice 1: $\quad \lim\limits_{\mathrm{x} \rightarrow 2} 4 \mathrm{x}-5=3$. What values of $\mathrm{x}$ guarantee that $\mathrm{f}(\mathrm{x})=4 \mathrm{x}-5$ is within

(a) 1 unit of 3?

(b) 0.08 units of 3?

(c) E units of 3? (Fig. 4)

The same ideas work even if the graphs of the functions are not straight lines, but the calculations are more complicated.

Example 2: $\quad \lim\limits_{\mathrm{x} \rightarrow 2} \mathrm{x}^{2}=4.$ (a) What values of $\mathrm{x}$ guarantee that $\mathrm{f}(\mathrm{x})=\mathrm{x}^{2}$ is within $\mathrm{1}$ unit of $\mathrm{4}$? (b) Within $0.2$ units of $4$?

(Fig. 5a) State each answer in the form "If $\mathrm{x}$ is within _____ units of $\mathrm{2}$, then $\mathrm{f}(\mathrm{x})$ is within $\mathrm{1}$ (or $0.2)$ unit of $\mathrm{4}$".

Solution; (a) If $x^{2}$ is within $\mathrm{1}$ unit of $\mathrm{4}$, then $3 < x^{2} < 5$ so $\sqrt{3} < x <$ $\sqrt{5}$

or $1.732 < \mathrm{x} < 2.236$. The interval containing these $\mathrm{x}$ values extends from $2-\sqrt{3} \approx 0.268$ units to the left of $\mathrm{2}$ to $\sqrt{5}-2 \approx 0.236$ units to the right of 2. Since we want to specify a single distance on each side of $\mathrm{2}$, we can pick the smaller of the two distances, $0.236$. (Fig. 5b)

"If $x$ is within $\mathrm{0.236}$ units of $\mathrm{2}$, then $\mathrm{f}(\mathrm{x})$ is within $\mathrm{1}$ unit of $\mathrm{4}$".

(b) Similarly, if $x^{2}$ is within $0.2$ units of 4, then $3.8 < x^{2} < 4.2$ so $\sqrt{3.8} < \mathrm{x} < \sqrt{4.2}$ or $1.949 < \mathrm{x} < 2.049$. The interval containing these $x$ values extends from $2-\sqrt{3.8} \approx 0.051$ units to the left of 2 to $\sqrt{4.2}-2 \approx 0.049$ units to the right of 2. Again picking the smaller of the two distances, "If $x$ is within $\mathrm{0.049}$ units of $\mathrm{2}$, then $\mathrm{f}(\mathrm{x})$ is within $\mathrm{1}$ unit of $\mathrm{4}$".

The situation in Example 2 of different distances on the left and right sides is very common, and we always pick our single distance to be the smaller of the distances to the left and right. By using the smaller distance, we can be certain that if $\mathrm{x}$ is within that smaller distance on either side, then the value of $\mathrm{f}(\mathrm{x})$ is within the specified distance of the value of the limit.

Practice 2: $\quad \lim\limits_{x \rightarrow 9} \sqrt{x}=3$. What values of $x$ guarantee that $\mathrm{f}(\mathrm{x})=\sqrt{\mathrm{x}}$ is within 1 unit of 3? Within $0.2$ units of 3? (Fig. 6) State each answer in the form.

"If $x$ is ______ within units of 2, then $\mathrm{f}(\mathrm{x})$ is within 1 (or 0.2) unit of 4".

The same ideas can also be used when the function and the specified distance are given graphically, and in that case we can give the answer graphically.

Example 3: In Fig. 7, $\lim\limits_{\mathrm{x} \rightarrow 2} \mathrm{f}(\mathrm{x})=3$. What values of $\mathrm{x}$ guarantee that $\mathrm{y}=\mathrm{f}(\mathrm{x})$ is within $\mathrm{E}$ units (given graphically) of 3? State your answer in the form "If $\mathrm{x}$ is within _____ (show a distance D graphically) of 2, then $\mathrm{f}(\mathrm{x})$ is within $\mathrm{E}$ units of 3".

Solution: $\quad$ The solution process requires several steps as illustrated in Fig. 8:

i. Use the given distance $\mathrm{E}$ to find the values $3-\mathrm{E}$ and $3+\mathrm{E}$ on the $\mathrm{y}$-axis.

ii. Sketch the horizontal band which has its lower edge at $\mathrm{y}=3-\mathrm{E}$ and its upper edge at $\mathrm{y}=3+\mathrm{E}$.

iii. Find the first locations to the right and left of $\mathrm{x}=2$ where the graph of $y=f(x)$ crosses the lines $y=3-E$ and $y=3+E$ and at these locations draw vertical lines to the $\mathrm{x}$-axis.

iv. On the $\mathrm{x}$-axis, graphically determine the distance from $\mathrm{2}$ to the vertical line on the left (labeled $\left.\mathrm{D}_{\mathrm{L}}\right)$ and from 2 to the vertical line on the right (labeled $\left.\mathrm{D}_{\mathrm{R}}\right)$.

v. Let the length $D$ be the smaller of the lengths $D_{L}$ and $D_{R}$.

Practice 3: In Fig. 9, $\lim\limits_{\mathrm{x} \rightarrow 3} \mathrm{f}(\mathrm{x})=1.8$. What of $x$ guarantee that $y=f(x)$ is within E units of $1.8$?

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 $\mathrm{g}$).

Practice 1: (a) $3-1 < 4 x-5 < 3+1$ so $7 < 4 x < 9$ and $1.75 < x < 2.25:$ " $x$ within $1 / 4$ unit of $\mathrm{2}$".

(b) $\quad 3-0.08 < 4 \mathrm{x}-5 < 3+0.08$ so $7.92 < 4 \mathrm{x} < 8.08$ and $1.98 < \mathrm{x} < 2.02: "\mathrm{x}$ within $0.02$ units of $\mathrm{2}$".

(c) $3-\mathrm{E} < 4 \mathrm{x}-5 < 3+\mathrm{E}$ so $8_­ \mathrm{E} < 4 \mathrm{x} < 8+\mathrm{E}$ and $2-\frac{\overline{\mathrm{E}}}{4} < \mathrm{x} < 2+{\mathrm{E}}{4}$: "x within $\mathrm{E} / 4$ units of $\mathrm{2}$".

Practice 2: "within 1 unit of 3": If $2 < \sqrt{x} < 4$, then $4 < x < 16$ which extends from 5 units to the left of 9 to 7 units to right of 9. Using the smaller of these two distances from 9, "If $x$ is within 5 units of 9, then $\sqrt{x}$ is within $\mathrm{1}$ unit of $\mathrm{3}$".

"within $0.2$ units of $3$": If $2.8 < \sqrt{x} < 3.2$, then $7.84 < x < 10.24$ which extends from $1.16$ units to the left of 9 to $1.24$ units to the right of 9. "If $x$ is within $1.16$ units of 9, then $\sqrt{x}$ is wqithin $0.2$ units of $\mathrm{3}$.

Practice 3: See Fig. 22.

Practice 4: See Fig. 23.

Practice 5: Given any $\varepsilon>0$, take $\delta=\varepsilon / 5$

If $x$ is within $\delta=\varepsilon / 5$ of 4, then

$\begin{array}{ll}-\varepsilon / 5 < \mathrm{x}-4 < \varepsilon / 5 & & \text { (subtracting 4) } \\ -\varepsilon < 5 \mathrm{x}-20 < \varepsilon & & \text { (multiplying by 5) } \\ -\varepsilon < (5 \mathrm{x}+3)-23 < \varepsilon & & \text { (rearranging to get the form we want)}\end{array}$

We have shown that "for any $\varepsilon > 0$, there is a $\delta > 0$ (namely $\delta=\varepsilon / 5)$" so that the rest of the definition is satisfied.

Practice 6: This is a much more sophisticated (= harder) problem.

Using "proof by contradiction" as outlined in the solution to Example 6. 

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

(ii) (See Fig. 24) Let $x_{1}$ be between $\mathrm{0}$ and $0+\delta$ and also require that $x_{1} < \frac{1}{2}$. Then $0 < \mathrm{x}_{1} < \frac{1}{2}$ so $\mathrm{f}\left(\mathrm{x}_{1}\right)=\frac{1}{\mathrm{x}_{1}} > 2$. Since $\mathrm{x}_{1}$ is within $\delta$ of $0, \mathrm{f}\left(\mathrm{x}_{1}\right) > 2$ is within $\varepsilon=1$ of $\mathrm{L}$, so L is greater than $2-\varepsilon=1: \mathbf{1} < \mathbf{L}$.

Let $\mathrm{x}_{2}$ be between 0 and $0-\delta$ and also require that $\mathrm{x}_{2} > -\frac{1}{2}$. Then $0 > \mathrm{x}_{2} > \frac{1}{2}$ so $\mathrm{f}\left(\mathrm{x}_{2}\right)=\frac{1}{\mathrm{x}_{2}} < -2$. Since $\mathrm{x}_{2}$ is within $\delta$ of $0, \mathrm{f}\left(\mathrm{x}_{2}\right) < -2$ is

$\varepsilon=1$ of $\mathrm{L}$, so $\mathrm{L}$ is less than $-2+\varepsilon=-1:-\mathbf{1} > \mathbf{L}$.

(iii) The two inequalities in bold print provide the contradiction we hoping to find, There is no value $\mathrm{L}$ that satisfies

BOTH $1 < \mathbf{L}$ and $\mathbf{L} < -\mathbf{1}$

so we can conclude that our assumption was false and that $\mathrm{f}(\mathrm{x})=\frac{1}{\mathrm{x}} \,$ does not have a limit as $\mathrm{x} \rightarrow 0$.