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# The Precise Definition of a Limit

### Learning Objectives

• Describe the epsilon-delta definition of a limit.
• Apply the epsilon-delta definition to find the limit of a function.
• Use the epsilon-delta definition to prove the limit laws.
• Use the epsilon-delta definition to find deltas algebraically.

By now you have progressed from the very informal definition of a limit in the introduction of this chapter to the intuitive understanding of a limit. At this point, you should have a very strong intuitive sense of what the limit of a function means and how you can find it. In this section, we convert this intuitive idea of a limit into a formal definition using precise mathematical language. The formal definition of a limit is quite possibly one of the most challenging definitions you will encounter early in your study of calculus; however, it is well worth any effort you make to reconcile it with your intuitive notion of a limit. Understanding this definition is the key that opens the door to a better understanding of calculus.

# Quantifying Closeness

Before stating the formal definition of a limit, we must introduce a few preliminary ideas. Recall that the distance between two points $a$ and $b$ on a number line is given by $|a-b|$.

• The statement $|f(x)-L|<\epsilon$ may be interpreted as: The distance between $f(x)$ and $L$ is less than $\epsilon$.
• The statement $0<|x-a|<\delta$ may be interpreted as: $x\ne a$ and the distance between $x$ and $a$ is less than $\delta$.

It is also important to look at the following equivalences for absolute value:

• The statement $|f(x)-L|<\epsilon$ is equivalent to the statement $L-\epsilon <f(x)<L+\epsilon$.
• The statement $0<|x-a|<\delta$ is equivalent to the statement $a-\delta <x<a+\delta$ and $x\ne a$.

With these clarifications, we can state the formal epsilon-delta definition of the limit.

### Definition

Let $f(x)$ be defined for all $x\ne a$ over an open interval containing $a$. Let $L$ be a real number. Then

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

if, for every $\epsilon >0$, there exists a $\delta >0$ such that if $0<|x-a|<\delta$, then $|f(x)-L|<\epsilon$.

This definition may seem rather complex from a mathematical point of view, but it becomes easier to understand if we break it down phrase by phrase. The statement itself involves something called a universal quantifier (for every $\epsilon >0$), an existential quantifier (there exists a $\delta >0$), and, last, a conditional statement (if $0<|x-a|<\delta$ then $|f(x)-L|<\epsilon$). Let’s take a look at (Figure), which breaks down the definition and translates each part.

Translation of the Epsilon-Delta Definition of the Limit
Definition Translation
1. For every $\epsilon >0$, 1. For every positive distance $\epsilon$ from $L$,
2. there exists a $\delta >0$, 2. There is a positive distance $\delta$ from $a$,
3. such that 3. such that
4. if $0<|x-a|<\delta$, then $|f(x)-L|<\epsilon$. 4. if $x$ is closer than $\delta$ to $a$ and $x\ne a$, then $f(x)$ is closer than $\epsilon$ to $L$.

We can get a better handle on this definition by looking at the definition geometrically. (Figure) shows possible values of $\delta$ for various choices of $\epsilon >0$ for a given function $f(x)$, a number $a$, and a limit $L$ at $a$. Notice that as we choose smaller values of $\epsilon$ (the distance between the function and the limit), we can always find a $\delta$ small enough so that if we have chosen an $x$ value within $\delta$ of $a$, then the value of $f(x)$ is within $\epsilon$ of the limit $L$.

Figure 1. These graphs show possible values of $\delta$, given successively smaller choices of $\epsilon$.

Visit the following applet to experiment with finding values of $\delta$ for selected values of $\epsilon$:

(Figure) shows how you can use this definition to prove a statement about the limit of a specific function at a specified value.

### Proving a Statement about the Limit of a Specific Function

Prove that $\underset{x\to 1}{\lim}(2x+1)=3$.

Let $\epsilon >0$.

The first part of the definition begins “For every $\epsilon >0$." This means we must prove that whatever follows is true no matter what positive value of $\epsilon$ is chosen. By stating “Let $\epsilon >0$," we signal our intent to do so.

Choose $\delta =\frac{\epsilon}{2}$.

The definition continues with “there exists a $\delta >0$.” The phrase “there exists” in a mathematical statement is always a signal for a scavenger hunt. In other words, we must go and find $\delta$. So, where exactly did $\delta =\epsilon/2$ come from? There are two basic approaches to tracking down $\delta$. One method is purely algebraic and the other is geometric.

We begin by tackling the problem from an algebraic point of view. Since ultimately we want $|(2x+1)-3|<\epsilon$, we begin by manipulating this expression: $|(2x+1)-3|<\epsilon$ is equivalent to $|2x-2|<\epsilon$, which in turn is equivalent to $|2||x-1|<\epsilon$. Last, this is equivalent to $|x-1|<\epsilon/2$. Thus, it would seem that $\delta =\epsilon/2$ is appropriate.

We may also find $\delta$ through geometric methods. (Figure) demonstrates how this is done.

Figure 2. This graph shows how we find $\delta$ geometrically.

Assume $0<|x-1|<\delta$. When $\delta$ has been chosen, our goal is to show that if $0<|x-1|<\delta$, then $|(2x+1)-3|<\epsilon$. To prove any statement of the form “If this, then that,” we begin by assuming “this” and trying to get “that.”

Thus,

[latex-display]\begin{array}{lllll}|(2x+1)-3| & =|2x-2| & & & \\ & =|2(x-1)| \\ & =|2||x-1| & & & \text{property of absolute values:} \, |ab|=|a||b| \\ & =2|x-1| & & & \\ & <2 \cdot \delta & & & \text{here’s where we use the assumption that} \, 0<|x-1|<\delta \\ & =2 \cdot \frac{\epsilon}{2}=\epsilon & & & \text{here’s where we use our choice of} \, \delta =\epsilon/2 \end{array}[/latex-display]

#### Analysis

In this part of the proof, we started with $|(2x+1)-3|$ and used our assumption $0<|x-1|<\delta$ in a key part of the chain of inequalities to get $|(2x+1)-3|$ to be less than $\epsilon$. We could just as easily have manipulated the assumed inequality $0<|x-1|<\delta$ to arrive at $|(2x+1)-3|<\epsilon$ as follows:

$\begin{array}{ll} 0<|x-1|<\delta & \implies |x-1|<\delta \\ & \implies -\delta <x-1<\delta \\ & \implies -\frac{\epsilon}{2}<x-1<\frac{\epsilon}{2} \\ & \implies -\epsilon <2x-2<\epsilon \\ & \implies |2x-2|<\epsilon \\ & \implies |(2x+1)-3|<\epsilon \end{array}$

Therefore, $\underset{x\to 1}{\lim}(2x+1)=3$. (Having completed the proof, we state what we have accomplished.)

After removing all the remarks, here is a final version of the proof:

Let $\epsilon >0$.

Choose $\delta =\epsilon/2$.

Assume $0<|x-1|<\delta$.

Thus,

$\begin{array}{ll} |(2x+1)-3|& =|2x-2| \\ & =|2(x-1)| \\ & =|2||x-1| \\ & =2|x-1| \\ & <2 \cdot \delta \\ & =2 \cdot \frac{\epsilon}{2} \\ & =\epsilon \end{array}$

Therefore, $\underset{x\to 1}{\lim}(2x+1)=3$.

The following Problem-Solving Strategy summarizes the type of proof we worked out in (Figure).

### Problem-Solving Strategy: Proving That $\underset{x\to a}{\lim}f(x)=L$ for a Specific Function $f(x)$

1. Let’s begin the proof with the following statement: Let $\epsilon >0$.
2. Next, we need to obtain a value for $\delta$. After we have obtained this value, we make the following statement, filling in the blank with our choice of $\delta$: Choose $\delta =$ _______.
3. The next statement in the proof should be (filling in our given value for $a$): Assume $0<|x-a|<\delta$.
4. Next, based on this assumption, we need to show that $|f(x)-L|<\epsilon$, where $f(x)$ and $L$ are our function $f(x)$ and our limit $L$. At some point, we need to use $0<|x-a|<\delta$.
5. We conclude our proof with the statement: Therefore, $\underset{x\to a}{\lim}f(x)=L$.

### Proving a Statement about a Limit

Complete the proof that $\underset{x\to -1}{\lim}(4x+1)=-3$ by filling in the blanks.

Let _____.

Choose $\delta =$ ________.

Assume $0<|x-\text{___}|<\delta$.

Thus, $|\text{________}-\text{___}| =|\text{_________}| = |\text{___}||\text{_________}| = \text{___} \, |\text{_______}| < \text{______} = \text{_______} = \epsilon$.

Therefore, $\underset{x \to -1}{\lim}(4x+1)=-3$.

We begin by filling in the blanks where the choices are specified by the definition. Thus, we have

Let $\epsilon >0$.

Choose $\delta =$ _______. (Leave this one blank for now -- we'll choose $\delta$ later)

Assume $0<|x-(-1)|<\delta$ (or equivalently, $0<|x+1|<\delta$).

Thus, $|(4x+1)-(-3)|=|4x+4|=|4||x+1|<4\delta = \text{_______} = \epsilon$.

Focusing on the final line of the proof, we see that we should choose $\delta =\frac{\epsilon}{4}$.

We now complete the final write-up of the proof:

Let $\epsilon >0$.

Choose $\delta =\frac{\epsilon}{4}$.

Assume $0<|x-(-1)|<\delta$ (or equivalently, $0<|x+1|<\delta$).

Thus, $|(4x+1)-(-3)|=|4x+4|=|4||x+1|<4\delta =4(\epsilon/4)=\epsilon$.

Complete the proof that $\underset{x\to 2}{\lim}(3x-2)=4$ by filling in the blanks.

Let _____.

Choose $\delta =$ ________.

Assume $0<|x-\text{___}|<\delta$.

Thus, $|\text{________}-\text{___}| =|\text{_________}| = |\text{___}||\text{_________}| = \text{___} \, |\text{_______}| < \text{______} = \text{_______} = \epsilon$.

Therefore, $\underset{x\to 2}{\lim}(3x-2)=4$.

Let $\epsilon >0$; choose $\delta =\frac{\epsilon}{3}$; assume $0<|x-2|<\delta$.

Thus, $|(3x-2)-4|=|3x-6|=|3| \cdot |x-2|<3 \cdot \delta =3 \cdot (\epsilon/3)=\epsilon$.

Therefore, $\underset{x\to 2}{\lim}3x-2=4$.

#### Hint

Follow the outline in the Problem-Solving Strategy that we worked out in full in (Figure).

In (Figure) and (Figure), the proofs were fairly straightforward, since the functions with which we were working were linear. In (Figure), we see how to modify the proof to accommodate a nonlinear function.

### Proving a Statement about the Limit of a Specific Function (Geometric Approach)

Prove that $\underset{x\to 2}{\lim}x^2=4$.

1. Let $\epsilon >0$. The first part of the definition begins “For every $\epsilon >0$," so we must prove that whatever follows is true no matter what positive value of $\epsilon$ is chosen. By stating “Let $\epsilon >0$," we signal our intent to do so.
2. Without loss of generality, assume $\epsilon \le 4$. Two questions present themselves: Why do we want $\epsilon \le 4$ and why is it okay to make this assumption? In answer to the first question: Later on, in the process of solving for $\delta$, we will discover that $\delta$ involves the quantity $\sqrt{4-\epsilon}$. Consequently, we need $\epsilon \le 4$. In answer to the second question: If we can find $\delta >0$ that “works” for $\epsilon \le 4$, then it will “work” for any $\epsilon >4$ as well. Keep in mind that, although it is always okay to put an upper bound on $\epsilon$, it is never okay to put a lower bound (other than zero) on $\epsilon$.
3. Choose $\delta =\text{min}\{2-\sqrt{4-\epsilon},\sqrt{4+\epsilon}-2\}$. (Figure) shows how we made this choice of $\delta$.
Figure 3. This graph shows how we find $\delta$ geometrically for a given $\epsilon$ for the proof in (Figure).
4. We must show: If $0<|x-2|<\delta$, then $|x^2-4|<\epsilon$, so we must begin by assuming
$0<|x-2|<\delta$.
We don’t really need $0<|x-2|$ (in other words, $x\ne 2$) for this proof. Since $0<|x-2|<\delta \implies |x-2|<\delta$, it is okay to drop $0<|x-2|$.
So, $|x-2|<\delta$, which implies $-\delta <x-2<\delta$.
Recall that $\delta =\text{min}\{2-\sqrt{4-\epsilon},\sqrt{4+\epsilon}-2\}$. Thus, $\delta \le 2-\sqrt{4-\epsilon}$ and consequently $-(2-\sqrt{4-\epsilon})\le -\delta$. We also use $\delta \le \sqrt{4+\epsilon}-2$ here. We might ask at this point: Why did we substitute $2-\sqrt{4-\epsilon}$ for $\delta$ on the left-hand side of the inequality and $\sqrt{4+\epsilon}-2$ on the right-hand side of the inequality? If we look at (Figure), we see that $2-\sqrt{4-\epsilon}$ corresponds to the distance on the left of 2 on the $x$-axis and $\sqrt{4+\epsilon}-2$ corresponds to the distance on the right. Thus,
$-(2-\sqrt{4-\epsilon})\le -\delta <x-2<\delta \le \sqrt{4+\epsilon}-2$.
We simplify the expression on the left:
$-2+\sqrt{4-\epsilon}<x-2<\sqrt{4+\epsilon}-2$.
Then, we add 2 to all parts of the inequality:
$\sqrt{4-\epsilon}<x<\sqrt{4+\epsilon}$.
We square all parts of the inequality. It is okay to do so, since all parts of the inequality are positive:
$4-\epsilon <x^2<4+\epsilon$.
We subtract 4 from all parts of the inequality:
$-\epsilon <x^2-4<\epsilon$.
Last,
$|x^2-4|<\epsilon$.
5. Therefore,
$\underset{x\to 2}{\lim}x^2=4$.

Find $\delta$ corresponding to $\epsilon >0$ for a proof that $\underset{x\to 9}{\lim}\sqrt{x}=3$.

Choose $\delta =\text{min}\{9-(3-\epsilon)^2,(3+\epsilon)^2-9\}$.

#### Hint

Draw a graph similar to the one in (Figure).

The geometric approach to proving that the limit of a function takes on a specific value works quite well for some functions. Also, the insight into the formal definition of the limit that this method provides is invaluable. However, we may also approach limit proofs from a purely algebraic point of view. In many cases, an algebraic approach may not only provide us with additional insight into the definition, it may prove to be simpler as well. Furthermore, an algebraic approach is the primary tool used in proofs of statements about limits. For (Figure), we take on a purely algebraic approach.

### Proving a Statement about the Limit of a Specific Function (Algebraic Approach)

Prove that $\underset{x\to -1}{\lim}(x^2-2x+3)=6$.

Let’s use our outline from the Problem-Solving Strategy:

1. Let $\epsilon >0$.
2. Choose $\delta =\text{min}\{1,\epsilon/5\}$. This choice of $\delta$ may appear odd at first glance, but it was obtained by taking a look at our ultimate desired inequality: $|(x^2-2x+3)-6|<\epsilon$. This inequality is equivalent to $|x+1|\cdot |x-3|<\epsilon$. At this point, the temptation simply to choose $\delta =\frac{\epsilon}{x-3}$ is very strong. Unfortunately, our choice of $\delta$ must depend on $\epsilon$ only and no other variable. If we can replace $|x-3|$ by a numerical value, our problem can be resolved. This is the place where assuming $\delta \le 1$ comes into play. The choice of $\delta \le 1$ here is arbitrary. We could have just as easily used any other positive number. In some proofs, greater care in this choice may be necessary. Now, since $\delta \le 1$ and $|x+1|<\delta \le 1$, we are able to show that $|x-3|<5$. Consequently, $|x+1| \cdot |x-3|<|x+1| \cdot 5$. At this point we realize that we also need $\delta \le \epsilon/5$. Thus, we choose $\delta =\text{min}\{1,\epsilon/5\}$.
3. Assume $0<|x+1|<\delta$. Thus,
$|x+1|<1$ and $|x+1|<\frac{\epsilon}{5}$.
Since $|x+1|<1$, we may conclude that $-1<x+1<1$. Thus, by subtracting 4 from all parts of the inequality, we obtain $-5<x-3<−1$. Consequently, $|x-3|<5$. This gives us
$|(x^2-2x+3)-6|=|x+1| \cdot |x-3|<\frac{\epsilon}{5} \cdot 5=\epsilon$.
Therefore,
$\underset{x\to -1}{\lim}(x^2-2x+3)=6$.

Complete the proof that $\underset{x\to 1}{\lim}x^2=1$.

Let $\epsilon >0$; choose $\delta =\text{min}\{1,\epsilon/3\}$; assume $0<|x-1|<\delta$.

Since $|x-1|<1$, we may conclude that $-1<x-1<1$. Thus, $1<x+1<3$. Hence, $|x+1|<3$.

$|x^2-1|=|x-1| \cdot |x+1|<\epsilon/3 \cdot 3=\epsilon$

#### Hint

Use (Figure) as a guide.

You will find that, in general, the more complex a function, the more likely it is that the algebraic approach is the easiest to apply. The algebraic approach is also more useful in proving statements about limits.

# Proving Limit Laws

We now demonstrate how to use the epsilon-delta definition of a limit to construct a rigorous proof of one of the limit laws. The triangle inequality is used at a key point of the proof, so we first review this key property of absolute value.

### Definition

The triangle inequality states that if $a$ and $b$ are any real numbers, then $|a+b|\le |a|+|b|$.

## Proof

We prove the following limit law: If $\underset{x\to a}{\lim}f(x)=L$ and $\underset{x\to a}{\lim}g(x)=M$, then $\underset{x\to a}{\lim}(f(x)+g(x))=L+M$.

Let $\epsilon >0$.

Choose $\delta_1>0$ so that if $0<|x-a|<\delta_1$, then $|f(x)-L|<\epsilon/2$.

Choose $\delta_2>0$ so that if $0<|x-a|<\delta_2$, then $|g(x)-M|<\epsilon/2$.

Choose $\delta =\text{min}\{\delta_1,\delta_2\}$.

Assume $0<|x-a|<\delta$.

Thus,

$0<|x-a|<\delta_1$ and $0<|x-a|<\delta_2$.

Hence,

$\begin{array}{ll} |(f(x)+g(x))-(L+M)| & =|(f(x)-L)+(g(x)-M)| \\ & \le |f(x)-L|+|g(x)-M| \\ & <\frac{\epsilon}{2}+\frac{\epsilon}{2}=\epsilon \end{array} _\blacksquare$

We now explore what it means for a limit not to exist. The limit $\underset{x\to a}{\lim}f(x)$ does not exist if there is no real number $L$ for which $\underset{x\to a}{\lim}f(x)=L$. Thus, for all real numbers $L$, $\underset{x\to a}{\lim}f(x)\ne L$. To understand what this means, we look at each part of the definition of $\underset{x\to a}{\lim}f(x)=L$ together with its opposite. A translation of the definition is given in (Figure).

Translation of the Definition of $\underset{x\to a}{\lim}f(x)=L$ and its Opposite
Definition Opposite
1. For every $\epsilon >0$, 1. There exists $\epsilon >0$ so that
2. there exists a $\delta >0$ so that 2. for every $\delta >0$,
3. if $0<|x-a|<\delta$, then $|f(x)-L|<\epsilon$. 3. There is an $x$ satisfying $0<|x-a|<\delta$ so that $|f(x)-L|\ge \epsilon$.

Finally, we may state what it means for a limit not to exist. The limit $\underset{x\to a}{\lim}f(x)$ does not exist if for every real number $L$, there exists a real number $\epsilon >0$ so that for all $\delta >0$, there is an $x$ satisfying $0<|x-a|<\delta$, so that $|f(x)-L|\ge \epsilon$. Let’s apply this in (Figure) to show that a limit does not exist.

### Showing That a Limit Does Not Exist

Show that $\underset{x\to 0}{\lim}\frac{|x|}{x}$ does not exist. The graph of $f(x)=\frac{|x|}{x}$ is shown here:

Suppose that $L$ is a candidate for a limit. Choose $\epsilon =1/2$.

Let $\delta >0$. Either $L\ge 0$ or $L<0$. If $L\ge 0$, then let $x=-\delta/2$. Thus,

$|x-0|=|-\frac{\delta}{2}-0|=\frac{\delta}{2}<\delta$

and

$|\frac{|-\frac{\delta}{2}|}{-\frac{\delta}{2}}-L|=|-1-L|=L+1\ge 1>\frac{1}{2}=\epsilon$.

On the other hand, if $L<0$, then let $x=\delta/2$. Thus,

$|x-0|=|\frac{\delta}{2}-0|=\frac{\delta}{2}<\delta$

and

$|\frac{|\frac{\delta}{2}|}{\frac{\delta}{2}}-L|=|1-L|=|L|+1\ge 1>\frac{1}{2}=\epsilon$.

Thus, for any value of $L$, $\underset{x\to 0}{\lim}\frac{|x|}{x}\ne L$.

# Finding Deltas Algebraically for Given Epsilons

Now that we have proven limits, we can now apply them with actual numbers for $\epsilon$ and $\delta$. Think of $\epsilon$ as the error in the x-direction and $\delta$ to be the error in the y-direction. These have applications in engineering when these errors are considered tolerances. We want to know what the error intervals are, and we are trying to minimize these errors.

### Finding deltas algebraically

Find an open interval about $x_0$ on which the inequality $|f(x)-L| < 0$ holds. Then give the largest value $\delta > 0$ such that for all $x$ satisfying $0 < |x-x_0| < \delta$ the inequality $|f(x)-L| < \epsilon$ holds.

[latex-display]f(x)=2x-8, L=6, x_0=7, \epsilon=0.14[/latex-display]

Answer: First we need will start with the inequality $|f(x)-L| < \epsilon$ and plug in our numbers. Then we will solve for x.$|2x-8 - 6| < \epsilon$$|2x-14| < \epsilon$$-0.14 < 2x - 14 < 0.14$$13.86 < 2x < 14.14$$6.93 < x < 7.07$Therefore, the interval is $(7.93,8.07)$For the second answer, we will start with $0 < |x-x_0| < \delta$.  We will plug in our value and solve:$|x-7| < \delta$$-\delta < x-7 < \delta$$7-\delta < x < 7+\delta$Now we will set each piece equal to the endpoints we found above.$7-\delta=7.93$ and $7+\delta=8.07$After solving we will get the same answer for each equation: $\delta=0.07$.

### Finding deltas algebraically

Find an open interval about $x_0$ on which the inequality $|f(x)-L| < 0$ holds. Then give the largest value $\delta > 0$ such that for all $x$ satisfying $0 < |x-x_0| < \delta$ the inequality $|f(x)-L| < \epsilon$ holds.

[latex-display]f(x)=\sqrt{x+4}, L=3, x_0=5, \epsilon=1[/latex-display]

Answer: First we need will start with the inequality $|f(x)-L| < \epsilon$ and plug in our numbers. Then we will solve for x.$|\sqrt(x+4) - 3| < \epsilon$$|\sqrt{x+4} - 3| < 1$$-1 < \sqrt{x+4} - 3 < 1$$2 < \sqrt{x+4} < 4$$4 < x + 4 < 16$$0 < x + 4 < 12$Therefore, the interval is $(0,12)$.For the second answer, we will start with $0 < |x-x_0| < \delta$.  We will plug in our value and solve:$|x-5| < \delta$$-\delta < x-5 < \delta$ [latex-display]5-\delta < x < 5+\delta[/latex-display] Now we will set each piece equal to the endpoints we found above. [latex-display]5-\delta=0[/latex] and $5+\delta=12[/latex-display] After solving we will get [latex]\delta=5 \text{ and }\delta=7$. Since the question is asking for the smallest interval, we choose the smaller number.  Therefore the answer is $\delta=5$.

### Key Concepts

• The intuitive notion of a limit may be converted into a rigorous mathematical definition known as the epsilon-delta definition of the limit.
• The epsilon-delta definition may be used to prove statements about limits.
• The epsilon-delta definition of a limit may be used to find deltas algebraically.

In the following exercises, write the appropriate $\epsilon$-$\delta$ definition for each of the given statements.

1. $\underset{x\to a}{\lim}f(x)=N$

2. $\underset{t\to b}{\lim}g(t)=M$

For every $\epsilon >0$, there exists a $\delta >0$ so that if $0<|t-b|<\delta$, then $|g(t)-M|<\epsilon$

3. $\underset{x\to c}{\lim}h(x)=L$

4. $\underset{x\to a}{\lim}\phi(x)=A$

For every $\epsilon >0$, there exists a $\delta >0$ so that if $0<|x-a|<\delta$, then $|\phi(x)-A|<\epsilon$

The following graph of the function $f$ satisfies $\underset{x\to 2}{\lim}f(x)=2$. In the following exercises, determine a value of $\delta >0$ that satisfies each statement.

5. If $0<|x-2|<\delta$, then $|f(x)-2|<1$.

6. If $0<|x-2|<\delta$, then $|f(x)-2|<0.5$.

$\delta \le 0.25$

The following graph of the function $f$ satisfies $\underset{x\to 3}{\lim}f(x)=-1$. In the following exercises, determine a value of $\delta >0$ that satisfies each statement.

7. If $0<|x-3|<\delta$, then $|f(x)+1|<1$.

8. If $0<|x-3|<\delta$, then $|f(x)+1|<2$.

$\delta \le 2$

The following graph of the function $f$ satisfies $\underset{x\to 3}{\lim}f(x)=2$. In the following exercises, for each value of $\epsilon$, find a value of $\delta >0$ such that the precise definition of limit holds true.

9. $\epsilon =1.5$

10. $\epsilon =3$

$\delta \le 1$

In the following exercises, use a graphing calculator to find a number $\delta$ such that the statements hold true.

11. [T] $|\sin (2x)-\frac{1}{2}|<0.1$, whenever $|x-\frac{\pi}{12}|<\delta$

12. [T] $|\sqrt{x-4}-2|<0.1$, whenever $|x-8|<\delta$

$\delta <0.3900$

In the following exercises, use the precise definition of limit to prove the given limits.

13. $\underset{x\to 2}{\lim}(5x+8)=18$

14. $\underset{x\to 3}{\lim}\frac{x^2-9}{x-3}=6$

Let $\delta =\epsilon$. If $0<|x-3|<\epsilon$, then $|x+3-6|=|x-3|<\epsilon$.

15. $\underset{x\to 2}{\lim}\frac{2x^2-3x-2}{x-2}=5$

16. $\underset{x\to 0}{\lim}x^4=0$

Let $\delta =\sqrt[4]{\epsilon}$. If $0<|x|<\sqrt[4]{\epsilon}$, then $|x^4|=x^4<\epsilon$.

17. $\underset{x\to 2}{\lim}(x^2+2x)=8$

In the following exercises, use the precise definition of limit to prove the given one-sided limits.

18. $\underset{x\to 5^-}{\lim}\sqrt{5-x}=0$

Let $\delta =\epsilon^2$. If $5-\epsilon^2<x<5$, then $|\sqrt{5-x}|=\sqrt{5-x}<\epsilon$.

19. $\underset{x\to 0^+}{\lim}f(x)=-2$, where $f(x)=\begin{cases} 8x-3 & \text{if} \, x<0 \\ 4x-2 & \text{if} \, x \ge 0 \end{cases}$.

20. $\underset{x\to 1^-}{\lim}f(x)=3$, where $f(x)=\begin{cases} 5x-2 & \text{if} \, x < 1 \\ 7x-1 & \text{if} x \ge 1 \end{cases}$

Let $\delta =\epsilon/5$. If $1-\epsilon/5<x<1$, then $|f(x)-3|=5x-5<\epsilon$.

In the following exercises, use the precise definition of limit to prove the given infinite limits.

21. $\underset{x\to 0}{\lim}\frac{1}{x^2}=\infty$

22. $\underset{x\to -1}{\lim}\frac{3}{(x+1)^2}=\infty$

Let $\delta =\sqrt{\frac{3}{N}}$. If $0<|x+1|<\sqrt{\frac{3}{N}}$, then $f(x)=\frac{3}{(x+1)^2}>N$.

23. $\underset{x\to 2}{\lim}-\frac{1}{(x-2)^2}=−\infty$

24. An engineer is using a machine to cut a flat square of Aerogel of area 144 cm2. If there is a maximum error tolerance in the area of 8 cm2, how accurately must the engineer cut on the side, assuming all sides have the same length? How do these numbers relate to $\delta, \, \epsilon, \, a$, and $L$?

0.033 cm, $\epsilon =8, \, \delta =0.33, \, a=12, \, L=144$

25. Use the precise definition of limit to prove that the following limit does not exist: $\underset{x\to 1}{\lim}\frac{|x-1|}{x-1}$.

26. Using precise definitions of limits, prove that $\underset{x\to 0}{\lim}f(x)$ does not exist, given that $f(x)$ is the ceiling function. (Hint: Try any $\delta <1$.)

27. Using precise definitions of limits, prove that $\underset{x\to 0}{\lim}f(x)$ does not exist: $f(x)=\begin{cases} 1 & \text{if} \, x \, \text{is rational} \\ 0 & \text{if} \, x \, \text{is irrational} \end{cases}$. (Hint: Think about how you can always choose a rational number $0<r<d$, but $|f(r)-0|=1$.)

28. Using precise definitions of limits, determine $\underset{x\to 0}{\lim}f(x)$ for $f(x)=\begin{cases} x & \text{if} \, x \, \text{is rational} \\ 0 & \text{if} \, x \, \text{is irrational} \end{cases}$ (Hint: Break into two cases, $x$ rational and $x$ irrational.)

0

29. Using the function from the previous exercise, use the precise definition of limits to show that $\underset{x\to a}{\lim}f(x)$ does not exist for $a\ne 0$.

For the following exercises, suppose that $\underset{x\to a}{\lim}f(x)=L$ and $\underset{x\to a}{\lim}g(x)=M$ both exist. Use the precise definition of limits to prove the following limit laws:

30. $\underset{x\to a}{\lim}(f(x)-g(x))=L-M$

$f(x)-g(x)=f(x)+(-1)g(x)$

31. $\underset{x\to a}{\lim}[cf(x)]=cL$ for any real constant $c$ (Hint: Consider two cases: $c=0$ and $c\ne 0$.)

32. $\underset{x\to a}{\lim}[f(x)g(x)]=LM$. (Hint: $|f(x)g(x)-LM|=|f(x)g(x)-f(x)M+f(x)M-LM|\le |f(x)||g(x)-M|+|M||f(x)-L|$.)

$\underset{x\to a}{\lim}f(x)=L$ if for every $\epsilon >0$, there exists a $\delta >0$ such that if $0<|x-a|<\delta$, then $|f(x)-L|<\epsilon$
If $a$ and $b$ are any real numbers, then $|a+b|\le |a|+|b|$