Mathematics 265 Introduction to Calculus I

Study Guide :: Unit 3

Limits

The Definition of a Limit

In earlier sections of this unit, we introduced the basic definitions and results we apply to evaluate limits. We provided some explanations to make them plausible, but it is one thing to accept something as plausible, and another to understand, without any reason for doubt, that something is true. Mathematicians look for logical arguments that explain why a statement or result is true. In this section, we consider how these results came about, so that your understanding of limit becomes deeper, and so that you can appreciate that all of the things we do with mathematics have a solid foundation. It is not our intention to prove all the results we have been using—time constraints prevent that. Instead, we will choose some of them, and we invite you to try to prove the others if your schedule allows it.

Understanding why something is true is what makes mathematics exciting. In our efforts to show that something is true, we must start with definitions and axioms.

We know that the statement $f\left(x\right)\to L$ as $x\to a$ means that we can make the values of $f\left(x\right)$ arbitrarily close to $L$ (as close as we like) by taking $x$ to be sufficiently close to $a$, but not equal to $a\text{.}$

This is the same as saying that we can make the values of $f\left(x\right)$ arbitrarily close to $L$ (as close as we like) because there are always $x$s sufficiently close to $a$ such that $f\left(x\right)$ is close to $L\text{.}$

Do you agree with this statement? Keep in mind the graph of a function with limit $L$ at $a$ (see Figure 3.29, below).

Figure 3.29. Function defined at points $a$ and $x$ close to $a$

In Figure 3.29, we can see that for a small number $m$ (as small as we like), we can find $x$ close to $a$, such that the distance between $f\left(x\right)$ and $L$ is less than $m\text{.}$ The challenge is to write precisely this idea using mathematical concepts; that is, to translate this idea into a mathematical statement.

We need to express the idea of “$x$ close to $a$” in mathematical terms. Two numbers are close if the distance between them is small, so we need to express the idea of “distance between two numbers.” What is the distance between the numbers $x$ and $a$? How do we measure the distance between $x$ and $a$?

Figure 3.30. Real number line showing $x<a$

If $x<a$, as shown in the real line above, then their distance is $a-x$, regardless of whether both are positive, both are negative, or one is positive and the other negative. Convince yourself of this fact. Figures 7 and 8 on page 341 of the textbook may help you.

Figure 3.31. Real number line showing $x>a$

If $x>a$, as shown in the real line above, then their distance is $x-a$, regardless of whether both are positive, both are negative, or one is positive and the other negative.

Combining these two results, we can say that the distance between two distinct numbers $x$ and $a$ is $|x-a|$ because

$\left|x-a\right|=\{\begin{array}{lll}x-a\hfill & \text{if}\hfill & x-a\ge 0\hfill \\ -\left(x-a\right)\hfill & \text{if}\hfill & x-a<0\hfill \end{array}=\{\begin{array}{lll}x-a\hfill & \text{if}\hfill & x\ge a\hfill \\ a-x\hfill & \text{if}\hfill & x<a\hfill \end{array}$

This is precisely what we say above!

As you can see, $|6-6.01|=0.01$ because the distance between $6$ and $6.01$ is $0.01$, and $|3.5-3|=0.5$, because the distance between $3$ and $3.5$ is $0.5\text{.}$

The distance between two distinct numbers $x$ and $a$ can be very small, and this fact can be expressed mathematically as follows:

$\left|x-a\right|<\delta \text{ for a small number}\delta >0\text{.}$

We use the Greek letter $\delta$ (delta) to denote a small number. It is a convention, and it has its historical reasons.

If we write $|x-3|<0.01$, then the distance between $x$ and $3$ is less than $0.01\text{.}$ These numbers are on the interval $\left(2.99,3.01\right)\text{.}$

Figure 3.32. Real number line showing the interval $\left(2.99,3.01\right)$

If we write $|x-3|<\delta$, then we are referring to the numbers $x$ whose distance to $3$ is less than the small number $\delta \text{.}$

When we write $0<|x-3|<\delta$, we are referring to all numbers $x$ very close to $3$, but not equal to $3$, because $|x-3|=0$ only of $x=3\text{.}$

Now we must express the idea of “making the values of $f\left(x\right)$ arbitrarily close to $L$ (as close as we like)” mathematically.

We know that “$f\left(x\right)$ is close to $L$” is written as $\left|f\left(x\right)-L\right|<\epsilon$ for a small number $\epsilon >0\text{.}$

Note that $\epsilon$ is the Greek letter epsilon.

But “the distance is as small as we want” is the same as saying that the distance is less than any number we want to give.

So, when we write

$\left|f\left(x\right)-L\right|<\epsilon \text{ for any given small number}\epsilon >0\text{,}$

we are saying that the distance between $f\left(x\right)$ and $L$ is as small as any number we want to give.

Hence, the statement

we can find a $\delta >0$ such that, for any $0<\left|x-a\right|<\delta \text{,}$ it is the case that $\left|f\left(x\right)-L\right|<\epsilon \text{.}$ for any given small number $\epsilon >0$

is saying that

we can find $x$s very close to $a$, but not equal to $a$, so that $f\left(x\right)$ can be as close to $L$ as we like.

This is precisely the definition of a limit.

Note that this definition expresses what we understood a limit to be.

We will need the following definition in our further discussions.

Definition 3.31 extends naturally to any finite number or numbers; for example,

${\displaystyle \text{min}\left\{3,-1,\frac{6}{2}\right\}=-1\text{and}\text{max}\left\{0,-\frac{1}{2},-\frac{1}{4}\right\}=0\text{.}}$

The results about limits that we presented to you are consequences of Definition 3.30 and the properties of the real numbers. For example, let us see why the first and third laws of limits in Theorem 3.23 are true.

Sum of limits: If $\underset{x\to a}{\lim }f\left(x\right)=L$ and $\underset{x\to a}{\lim }g\left(x\right)=M$, then

$\underset{x\to a}{\lim }f\left(x\right)+g\left(x\right)=L+M\text{.}$

Proof. We start with knowing what we want to show (prove).

Given any $\epsilon >0$, we can find $\delta >0$ such that, for any $0<|x-a|<\delta$, it is the case that

$|f\left(x\right)+g\left(x\right)-\left(L+M\right)|<\epsilon \text{.}$

We are given $\epsilon >0$, and we must find $\delta >0$ with the property indicated above.

To proceed from this point, we must determine what we know.

In this case, we know that $\underset{x\to a}{\lim }f\left(x\right)=L$ and $\underset{x\to a}{\lim }g\left(x\right)=M\text{.}$ Definition 3.30 is true for any positive number, so

1. for ${\displaystyle \frac{\epsilon }{2}>0}$, there is a ${\delta }_1>0$ such that, for any $0<\left|x-a\right|<{\delta }_1\text{,}$ it is the case that ${\displaystyle \left|f\left(x\right)-L\right|<\frac{\epsilon }{2}\text{,}}$ and
2. for ${\displaystyle \frac{\epsilon }{2}>0}$, there is a ${\delta }_2>0$ such that, for any $0<\left|x-a\right|<{\delta }_2\text{,}$ it is the case that ${\displaystyle \left|g\left(x\right)-M\right|<\frac{\epsilon }{2}\text{.}}$

If $\delta =min\left({\delta }_1,{\delta }_2\right)$, then $\delta \le {\delta }_1$ and $\delta \le {\delta }_2\text{.}$ So, for any $x$ such that $0<\left|x-a\right|<\delta \text{,}$ it is true that $0<\left|x-a\right|<{\delta }_1$ and $0<\left|x-a\right|<{\delta }_2\text{.}$

Hence, statements 1 and 2 above are true for this $\delta \text{.}$

If $0<\left|x-a\right|<\delta \text{,}$ then ${\displaystyle \left|f\left(x\right)-L\right|<\frac{\epsilon }{2}\text{, and }\left|g\left(x\right)-M\right|<\frac{\epsilon }{2}\text{.}}$

From the triangle inequality, we have

${\displaystyle \left|f\left(x\right)+g\left(x\right)-\left(L+M\right)\right|\le \left|f\left(x\right)-L\right|+\left|g\left(x\right)-M)\right|<\frac{\epsilon }{2}+\frac{\epsilon }{2}=\epsilon \text{.}}$

Therefore, given $\epsilon >0$, we have found $\delta =min\left({\delta }_1,{\delta }_2\right)$, such that if $0<\left|x-a\right|<\delta$ then $\left|f\left(x\right)+g\left(x\right)-\left(L+M\right)\right|<\epsilon \text{.}$

Q.E.D

Note: The end of a proof is indicated by the initials “Q.E.D.,” an abbreviation of the Latin phrase quod erat demonstrandum (“which was to be proved”).

Product of limits: If $\underset{x\to a}{\lim }f\left(x\right)=L$ and $\underset{x\to a}{\lim }g\left(x\right)=M$, then

$\underset{x\to a}{\lim }f\left(x\right)g\left(x\right)=LM\text{.}$

Proof. We want to show that, for any given $\epsilon >0$, we can find a $\delta >0$ such that

$\text{for any }0<\left|x-a\right|<\delta \text{ it is the case that }\left|f\left(x\right)g\left(x\right)-LM\right|<\epsilon$

We know that $\underset{x\to a}{\lim }f\left(x\right)=L$ and $\underset{x\to a}{\lim }g\left(x\right)=M$; hence,

1. for any ${\epsilon }_1>0$, there is a ${\delta }_1>0$ such that for any $0<\left|x-a\right|<{\delta }_1\text{,}$ it is the case that $\left|f\left(x\right)-L\right|<{\epsilon }_1\text{,}$ and
2. for any ${\epsilon }_2>0$, there is a ${\delta }_2>0$ such that for any $0<\left|x-a\right|<{\delta }_2\text{,}$ it is the case that $\left|g\left(x\right)-M\right|<{\epsilon }_2\text{.}$

See how the addition of $0=g\left(x\right)L-g\left(x\right)L$ allows us to relate what we want to show with what we know.

By the triangle inequality,

$\begin{array}{ll}\left|f\left(x\right)g\left(x\right)-LM\right|\hfill & =|f\left(x\right)g\left(x\right)-LM+g\left(x\right)L-g\left(x\right)L|\hfill \\ \hfill & =\left|f\left(x\right)g\left(x\right)-g\left(x\right)L+g\left(x\right)L-LM\right|\hfill \\ \hfill & =|g\left(x\right)\left(f\left(x\right)-L\right)+L\left(g\left(x\right)-M\right)|\hfill \\ \hfill & \le \left|g\left(x\right)\right|\left|f\left(x\right)-L\right|+\left|L\right|\left|\left(g\left(x\right)-M\right)\right|\hfill \end{array}$

We want the final inequality above to be less than any $\epsilon >0$, and we can achieve this result by taking particular values for ${\epsilon }_1>0$ and ${\epsilon }_2>0$ in statements 1 and 2 above.

In statement 1, we take ${\displaystyle {\epsilon }_1=\frac{\epsilon }{2\left(1+\left|M\right|\right)}>0}\text{.}$ So, there is a ${\delta }_1>0$ such that

3. ${\displaystyle \left|f\left(x\right)-L\right|<\frac{\epsilon }{2\left(1+\left|M\right|\right)}\text{ for any }0<\left|x-a\right|<{\delta }_1\text{.}}$

In statement 2 we take ${\epsilon }_2=1\text{.}$ So, for $1>0$ there is a ${\delta }_2>0$ such that, for any $0<\left|x-a\right|<{\delta }_2\text{,}$ it is the case that $\left|g\left(x\right)-M\right|<1.$

Since $\left|g\left(x\right)\right|=\left|g\left(x\right)-M+M\right|\le \left|g\left(x\right)-M\right|+\left|M\right|\text{,}$ we conclude that

4. $\left|g\left(x\right)\right|\le \text{1+}\left|M\right|\text{ for any }0<\left|x-a\right|<{\delta }_2\text{.}$

Again, in statement 2, we take ${\displaystyle {\epsilon }_3=\frac{\epsilon }{2\left|L\right|}>0}\text{.}$ So, there is a ${\delta }_3>0$ such that

5. ${\displaystyle \left|g\left(x\right)-M\right|<\frac{\epsilon }{2\left|L\right|}\text{, for any }0<\left|x-a\right|<{\delta }_3\text{.}}$

Hence, if $\delta =min\left({\delta }_1,{\delta }_2,{\delta }_3\right)$, then, for any $0<|x-a|<\delta$, statements 3, 4 and 5, above, are valid. Therefore,

${\displaystyle \begin{array}{ll}\left|f\left(x\right)g\left(x\right)-LM\right|\hfill & \le \left|g\left(x\right)\right|\left|f\left(x\right)-L\right|+\left|L\right|\left|\left(g\left(x\right)-M\right)\right|\hfill \\ \hfill & <\left(1+\left|M\right|\right)\frac{\epsilon }{2\left(1+\left|M\right|\right)}+\left|L\right|\frac{\epsilon }{2\left|L\right|}=\epsilon \text{.}\hfill \end{array}}$

Q.E.D

Limit of the reciprocal: If $\underset{x\to a}{\lim }g\left(x\right)=M$ and $M\ne 0$, then ${\displaystyle \underset{x\to a}{\lim }\frac{1}{g\left(x\right)}=\frac{1}{M}\text{.}}$

Proof. We want to show that for any given $\epsilon >0$, we can find a $\delta >0$ such that for any

${\displaystyle 0<\left|x-a\right|<\delta \text{ it is the case that }\left|\frac{1}{g\left(x\right)}-\frac{1}{M}\right|<\epsilon \text{.}}$

We know that $\underset{x\to a}{\lim }g\left(x\right)=M$; hence,

1. for any ${\epsilon }_1>0$, there is a ${\delta }_1>0$ such that for any $0<\left|x-a\right|<{\delta }_1$ it is the case that $\left|g\left(x\right)-M\right|<{\epsilon }_1\text{.}$

To relate what we know to what we want to show, we do the operations and obtain

${\displaystyle \left|\frac{1}{g\left(x\right)}-\frac{1}{M}\right|=\left|\frac{M-g\left(x\right)}{g\left(x\right)M}\right|=\left|g\left(x\right)-M\right|\frac{1}{\left|g\left(x\right)M\right|}}$

From statement 1, if $\left|g\left(x\right)-M\right|<{\epsilon }_1\text{,}$ then by the properties of the absolute value,

${\epsilon }_1>\left|g\left(x\right)-M\right|=\left|M-g\left(x\right)\right|\ge \left|M\right|-\left|g\left(x\right)\right|\text{.}$

Therefore,

${\displaystyle \left|g\left(x\right)\right|>\left|M\right|-{\epsilon }_1\text{ and }\frac{1}{\left|Mg\left(x\right)\right|}<\frac{1}{\left|M\right|\left(\left|M\right|-{\epsilon }_1\right)}\text{.}}$

Hence,

${\displaystyle \left|g\left(x\right)-M\right|\frac{1}{\left|g\left(x\right)M\right|}<\frac{{\epsilon }_1}{\left|M\right|\left(\left|M\right|-{\epsilon }_1\right)}\text{.}}$

If we want ${\displaystyle \epsilon =\frac{{\epsilon }_1}{\left|M\right|\left(\left|M\right|-{\epsilon }_1\right)}\text{,}}$ we solve for ${\epsilon }_1$, and we determine that

${\displaystyle {\epsilon }_1=\frac{\epsilon {\left|M\right|}^2}{1+\left|M\right|\epsilon }>0.}$

From statement 1, above, we know that for this value of ${\epsilon }_1$, there is a $\delta$ such that if

${\displaystyle 0<\left|x-a\right|<\delta \text{ then }\left|g\left(x\right)-M\right|<\frac{\epsilon {\left|M\right|}^2}{1+\left|M\right|\epsilon }\text{.}}$

We then conclude that

${\displaystyle \left|\frac{1}{g\left(x\right)}-\frac{1}{M}\right|=\left|g\left(x\right)-M\right|\frac{1}{\left|g\left(x\right)M\right|}<\frac{{\epsilon }_1}{\left|M\right|\left(\left|M\right|-{\epsilon }_1\right)}=\epsilon \text{.}}$

Finally, we can use the fact that

${\displaystyle \frac{f\left(x\right)}{g\left(x\right)}=f\left(x\right)\frac{1}{g\left(x\right)}\text{,}}$

and determine, from the product of limits and the limit of the reciprocal, the limit of the quotient.

If $\underset{x\to a}{\lim }f\left(x\right)=L$ and $\underset{x\to a}{\lim }g\left(x\right)=M$ where $M\ne 0$, then

${\displaystyle \underset{x\to a}{\lim }\frac{f\left(x\right)}{g\left(x\right)}=\underset{x\to a}{\lim }f\left(x\right)\frac{1}{g\left(x\right)}=\underset{x\to a}{\lim }f\left(x\right)\underset{x\to a}{\lim }\frac{1}{g\left(x\right)}=L\frac{1}{M}=\frac{L}{M}}$

The formal definition of the side limits should indicate when an $x$ is very close to $a$ from the right (or left) but is not $a\text{.}$ We denote this situation by

$0<x-a<\delta \left(\text{or}0<a-x<\delta \right)\text{.}$

Observe that the statement $0<x-a$ says that $x\ne a$ and that $a<x$; that is, $x$ is on the right of $a\text{.}$

Definition 3.32.

1. Let $f$ be a function defined on the right of $a\text{.}$

   The limit of $f$, as $x$ approaches $a$ from the right, is $L$ if,

   given any $\epsilon >0$, we can find $\delta >0$ such that, for any $0<x-a<\delta$, we have $\left|f\left(x\right)-L\right|<\epsilon \text{.}$

2. Let $f$ be a function defined on the left of $a\text{.}$

   The limit of $f$, as $x$ approaches $a$ from the left, is $L$ if,

   given any $\epsilon >0$, we can find $\delta >0$ such that, for any $0<a-x<\delta$, we have $\left|f\left(x\right)-L\right|<\epsilon \text{.}$

Remember that a function $f$ is continuous at $a$ if, $\underset{x\to a}{\lim }f\left(x\right)=f\left(a\right)\text{.}$ By Definition 3.30, the formal definition of continuity is as follows.

From the laws of limits we have the following laws for continuous functions.

Proposition 3.34. If $f$ and $g$ are continuous at $a$, then

1. $\underset{x\to a}{\lim }f\left(x\right)\pm g\left(x\right)=f\left(a\right)\pm g\left(a\right)\text{.}$
2. $\underset{x\to a}{\lim }cf\left(x\right)=cf\left(a\right)$, for any constant $c\text{.}$
3. $\underset{x\to a}{\lim }f\left(x\right)g\left(x\right)=f\left(a\right)g\left(a\right)\text{.}$
4. ${\displaystyle \underset{x\to a}{\lim }\frac{f\left(x\right)}{g\left(x\right)}=\frac{f\left(a\right)}{g\left(a\right)}}$, if $g\left(a\right)\ne 0\text{.}$

Proposition 3.34 is saying that the sum (difference), constant product, product and quotient of continuous functions at $a$ are continuous at $a\text{.}$

Example 3.78. Using Definition 3.33, above, we can also prove that the identity function $I\left(x\right)=x$ is continuous everywhere. We want to show that, for any number $a$, the function is continuous at $a$; that is, given $\epsilon >0$, we can find $\delta >0$ such that

for any $0<\left|x-a\right|<\delta \text{,}$ it is the case that $\left|I\left(x\right)-I\left(a\right)\right|<\epsilon \text{.}$

Since $I\left(x\right)=x$, we have $\left|I\left(x\right)-I\left(a\right)\right|\text{=}\left|x-a\right|\text{,}$ and for $\epsilon =\delta$, the statement is true. Since $a$ is any number, the function $I$ is continuous everywhere.

Example 3.79. By Example 3.78 and Proposition 3.34(c), we know that for any positive integer $n$, the function $I{\left(x\right)}^n=x^n$ is the $n$ product of the everywhere-continuous identity function. Hence, by the second law of limits, the function $f\left(x\right)=c{x}^n$ is continuous everywhere for any constant $c\text{.}$ You can conclude, again by the law of limits, that polynomial functions are continuous everywhere.

Example 3.80. Since a rational function is the quotient of polynomial functions, rational functions are continuous on their domains by Example 3.79 and Proposition 3.34(d).

Example 3.81. By the definition of radians, for any numbers (radians) $\theta$ and $\eta$, it is true that $\left|\sin \theta -\text{ sin }\eta \right|\le \left|\theta -\eta \right|\text{.}$

Let us take any $\epsilon >0\text{.}$ Then, for $\epsilon =\delta$, we have that, for $\left|\theta -\eta \right|<\delta \text{,}$ it is true that $\left|\sin \theta -\sin \eta \right|<\epsilon \text{.}$

This result shows that the sine function is continuous at $\eta$; however, since $\eta$ is arbitrary, we conclude that sine is continuous everywhere.

Theorem 3.35. If the function $g$ is continuous at $a$, and the function $f$ is continuous at $g\left(a\right)$, then the composition $f\circ g$ is continuous at $a\text{.}$ [Note that this theorem was presented earlier as Theorem 3.13.]

Proof. We have $\underset{x\to a}{\lim }g\left(x\right)=g\left(a\right)$ and $\underset{x\to b}{\lim }f\left(x\right)=f\left(b\right)$ for $b=g\left(a\right)\text{.}$ Then, by Exercise 34(b), which you just completed, $\underset{x\to a}{\lim }f\left(g\left(x\right)\right)=f\left(g\left(a\right)\right)$, and therefore, $\underset{x\to a}{\lim }f\left(g\left(x\right)\right)$ is continuous at $a\text{.}$

Q.E.D

Example 3.82. From the addition formula of the sine function, we have

${\displaystyle \cos x=\sin \left(x+\frac{\pi }{2}\right)\text{.}}$

Hence, cosine is the composition of the outside function sine and the inside polynomial function $x+\pi ∕2\text{.}$

By Examples 3.79 and 3.81, these two functions are continuous everywhere; hence, by Theorem 3.35, cosine is continuous everywhere.

We conclude from Examples 3.79 to 3.82 and Proposition 3.34 that Theorem 3.14 is true.

At this point, we want to consider the formal definition of

$\underset{x\to a^{+}}{\lim }f\left(x\right)=\pm \infty ,\underset{x\to a^{-}}{\lim }f\left(x\right)=\pm \infty \text{and}\underset{x\to a}{\lim }f\left(x\right)=\pm \infty \text{.}$

By Definition 3.6, we have $\underset{x\to a^{+}}{\lim }f\left(x\right)=\infty$ if we can make the values of $f\left(x\right)$ arbitrarily large (as large as we like) by taking $x$ to be sufficiently close to $a$ from the right but not equal to $a\text{.}$

This statement is the same as saying that it does not matter how large a number $M$ we have, we can always find an $x$ close to $a$ from the right, but not equal to $a$, such that $f\left(x\right)$ is bigger than $M\text{.}$

Do you agree with this statement? Again, keep in mind the graph of a function with an infinite limit at $a$ from the right.

Figure 3.33. Function with an infinite positive limit at $a$ from the right

From Figure 3.33, above, we can see that, given any large $M>0$, we can find $\delta >0$ such that, for any $0<x-a<\delta$, we have $f\left(x\right)>M\text{.}$

Figure 3.34. Function with an infinite negative limit at $a$ from the right

Figure 3.34, above, is the graph of a function with a negative infinite limit at $a$ from the right. We can see that it does not matter how small and negative a number $N$ we have, we can always find an $x$ close to $a$ from the right, but not equal to $a$, so that $f\left(x\right)$ is less than $N\text{.}$

In Figure 3.34, we can see that, given a small $N<0$ we can find $\delta >0$ such that for any $0<x-a<\delta$, we have $f\left(x\right)<N\text{.}$

We leave it to you to analyse the limits $\underset{x\to a^{-}}{\lim }f\left(x\right)=\pm \infty$ and $\underset{x\to a}{\lim }f\left(x\right)=\pm \infty$ in the definition below.

In the discussion below, we present some proofs about infinite limits. We strongly recommend that you try other proofs.

Theorem 3.38. If $\underset{x\to a^{+}}{\lim }g\left(x\right)=0$ and $g\left(x\right)>0$ for $x\to a^{+}$, then

${\displaystyle \underset{x\to a^{+}}{\lim }\frac{1}{g\left(x\right)}=\infty \text{.}}$

Proof. We want to show that, given $M>0$, there is a $\delta >0$ such that, for any $0<x-a<\delta$, we have ${\displaystyle \frac{1}{g\left(x\right)}>M\text{.}}$

We have

1. for any $\epsilon >0$, there is ${\delta }_1>0$ such that $\left|g\left(x\right)\right|<\epsilon$ for any $0<x-a<{\delta }_1\text{.}$
2. there is a ${\delta }_2>0$, such that $g\left(x\right)>0$ for any $0<x-a<{\delta }_2\text{.}$

In particular, for ${\displaystyle \epsilon =\frac{1}{M}>0}$ in statement 1, we have

3. there is ${\delta }_1>0$ such that ${\displaystyle \left|g\left(x\right)\right|<\frac{1}{M}}$ for any $0<x-a<{\delta }_1\text{.}$

If $\delta =min\left({\delta }_1,{\delta }_2\right)$, then for $0<x-a<\delta$, statements 2 and 3 hold, and

${\displaystyle g\left(x\right)>0\text{ and }\left|g\left(x\right)\right|<\frac{1}{M}\text{.}}$

Hence,

${\displaystyle \left|g\left(x\right)\right|\text{=g}\left(x\right)\text{ and }\frac{1}{g\left(x\right)}>\text{M}\text{.}}$

Q.E.D

Theorem 3.39. If $\underset{x\to a}{\lim }f\left(x\right)=L>0$, and $\underset{x\to a}{\lim }g\left(x\right)=\infty$, then

$\underset{x\to a}{\lim }f\left(x\right)g\left(x\right)=\infty \text{.}$

Proof. We want to show that, given $M>0$, there is a $\delta >0$ such that, for any $0<\left|x-a\right|<\delta \text{,}$ we have $f\left(x\right)g\left(x\right)>M\text{.}$

We have

1. for any $\epsilon >0$, there is ${\delta }_1>0$ such that $\left|f\left(x\right)-L\right|<\epsilon$ for any $0<\left|x-a\right|<{\delta }_1\text{.}$
2. for any $M^{\prime }>0$, there is a ${\delta }_2>0$ such that $g\left(x\right)>M^{\prime }$ for any $0<\left|x-a\right|<{\delta }_2\text{.}$

In particular, in statement 1, for

${\displaystyle \epsilon =\frac{L}{2}>0\text{,}}$

there is ${\delta }_1>0$ such that

${\displaystyle \left|f\left(x\right)-L\right|<\frac{L}{2}\text{ for any }0<\left|x-a\right|<{\delta }_1\text{.}}$

Hence,

${\displaystyle -\frac{L}{2}<f\left(x\right)-L<\frac{L}{2}\text{.}}$

Adding $L$ to each side of the inequality, we conclude that

3. ${\displaystyle \frac{L}{2}<f\left(x\right)<\frac{3L}{2}\text{ for }0<\left|x-a\right|<{\delta }_1\text{.}}$

In statement 2, we take ${\displaystyle M^{\prime }=\frac{2M}{L}>0}\text{.}$ There is a ${\delta }_2>0$, such that

4. ${\displaystyle g\left(x\right)>\frac{2M}{L}\text{ for any }0<\left|x-a\right|<{\delta }_2\text{.}}$

If $\delta =min\left({\delta }_1,{\delta }_2\right)$, then statements 3 and 4 hold for all $0<\left|x-a\right|<\delta \text{,}$ and we have

${\displaystyle f\left(x\right)g\left(x\right)>\frac{L}{2}\frac{2M}{L}=M\text{.}}$

Q.E.D

Finally, let us consider the formal definition of

$\underset{x\to \pm \infty }{\lim }f\left(x\right)=L\text{.}$

The limit

$\underset{x\to \infty }{\lim }f\left(x\right)=L$

means that the values of $f\left(x\right)$ can be made arbitrarily close to $L$ by taking $x$ sufficiently large and positive (see Figure 3.35, below).

Figure 3.35. Formal definition of $\underset{x\to +\infty }{\lim }f\left(x\right)=L$

For any small number $\epsilon >0$ (as small as we want), we can always find a large number $M>0$ such that, for any $x>M$, the distance between $f\left(x\right)$ and $L$ is less than $\epsilon \text{.}$

We leave it to you to give the formal definitions of the statements in Definition 3.40 for $x\to a^{-}$ and $x\to a\text{.}$

The limit

$\underset{x\to -\infty }{\lim }f\left(x\right)=L$

means that the values of $f\left(x\right)$ can be made arbitrarily close to $L$ by taking $x$ sufficiently small and negative (see Figure 3.36, below).

For any small number $\epsilon >0$ (as small as we want), we can always find a small number $N<0$ such that, for any $x<N$, the distance between $f\left(x\right)$ and $L$ is less than $\epsilon \text{.}$

Figure 3.36. Formal definition of $\underset{x\to -\infty }{\lim }f\left(x\right)=L$

We have provided a set of theorems that hold for limits at infinity; one of them is the laws of limits. Compare the following proof with the product of limits we gave before (see Theorem 3.23).

Theorem 3.42. If $\underset{x\to \infty }{\lim }f\left(x\right)=L$ and $\underset{x\to \infty }{\lim }g\left(x\right)=K$, then

$\underset{x\to \infty }{\lim }f\left(x\right)g\left(x\right)=LK\text{.}$

Proof. We want to show that, for any given $\epsilon >0$, we can find an $M>0$ such that, for any

$x>M$ it is the case that $\left|f\left(x\right)g\left(x\right)-LM\right|<\epsilon \text{.}$

We know that, for any positive number,

1. ${\epsilon }_1>0$, there is an $M_1>0$ such that, for any $x>M_1$, it is the case that $\left|f\left(x\right)-L\right|<{\epsilon }_1\text{.}$
2. ${\epsilon }_2>0$, there is an $M_2>0$ such that, for any $x>M_2$, it is the case that $\left|g\left(x\right)-K\right|<{\epsilon }_2\text{.}$

As before, we add $0=g\left(x\right)L-g\left(x\right)L$ to relate what we want to show with what we know. By the triangle inequality, we know that

$\begin{array}{ll}\left|f\left(x\right)g\left(x\right)-LK\right|\hfill & =\left|f\left(x\right)g\left(x\right)-g\left(x\right)L+g\left(x\right)L-LK\right|\hfill \\ \hfill & \le \left|g\left(x\right)\right|\left|f\left(x\right)-L\right|+\left|L\right|\left|g\left(x\right)-K\right|\hfill \end{array}$

Since we want this inequality to be less than $\epsilon$, and since statements 1 and 2 above are true for any positive real number, we take values for ${\epsilon }_1$ and ${\epsilon }_2$ that fit these criteria.

In statement 2, we take ${\epsilon }_2=1$, and we find that for $1>0$, there is an $M_2>0$ such that for any $x>M_2$ it is the case that $\left|g\left(x\right)-K\right|<{\epsilon }_2\text{.}$

Since $\left|g\left(x\right)\right|=\left|g\left(x\right)-K+K\right|\le \left|g\left(x\right)-K\right|+\left|K\right|\text{,}$ we conclude that

3. $\left|g\left(x\right)\right|<1+\left|K\right|$ for any $x>M_2\text{.}$

Again, in statement 2, we take ${\displaystyle {\epsilon }_2=\frac{\epsilon }{2\left|L\right|}>0\text{,}}$ and we find that there is an $M_3>0$ such that

4. ${\displaystyle \left|g\left(x\right)-K\right|<\frac{\epsilon }{2\left|L\right|}}$ for any $x>M_3\text{.}$

In statement 1, we take ${\displaystyle {\epsilon }_1=\frac{\epsilon }{2\left(1+\left|K\right|\right)}>0\text{,}}$ and we find that there is an $M_1>0$ such that

5. ${\displaystyle \left|f\left(x\right)-L\right|<\frac{\epsilon }{2\left(1+\left|K\right|\right)}}$ for any $x>M_1\text{.}$

Hence, if $M=max\left(M_1,M_2,M_3\right)$, then for any $x>M$, statements 3, 4 and 5 are valid. Therefore,

${\displaystyle \begin{array}{ll}\left|f\left(x\right)g\left(x\right)-LK\right|\hfill & \le \left|g\left(x\right)\right|\left|f\left(x\right)-L\right|+\left|L\right|\left|g\left(x\right)-K\right|\hfill \\ \hfill & <\left(1+\left|K\right|\right)\frac{\epsilon }{2\left(1+\left|K\right|\right)}+\left|L\right|\frac{\epsilon }{2\left|L\right|}=\epsilon \hfill \end{array}}$

Q.E.D

From this proof, you can see intuitively why some results for limits at finite numbers are also true for limits at infinity.

Next, we give some other proofs to illustrate the different arguments we can use to prove a statement logically. Observe also how we use the properties of the absolute value.

Theorem 3.43. If $\underset{x\to a}{\lim }f\left(x\right)=L$ and $\underset{x\to a}{\lim }g\left(x\right)=K$ and $g\left(x\right)\le f\left(x\right)$ for $x\to a$, then $K\le L\text{.}$

This theorem is saying that if $g\left(x\right)\le f\left(x\right)$ for $x\to a$, then $\underset{x\to a}{\lim }g\left(x\right)\le \underset{x\to a}{\lim }f\left(x\right)\text{.}$

Proof. We will argue by contradiction. If we assume that $K>L$, this assumption will take us to a contradiction, and from the contradiction, we can conclude that our assumption is false, and that $K\le L\text{.}$

By the sum of limits, we know that

$\underset{x\to a}{\lim }g\left(x\right)-f\left(x\right)=K-L\text{.}$

If $K>L$, then $K-L>0\text{.}$ We have two hypotheses:

1. for any ${\epsilon }_1>0$, there is a ${\delta }_1>0$ such that $\left|g\left(x\right)-K\right|<{\epsilon }_1$ for any $0<\left|x-a\right|<{\delta }_1\text{.}$
2. for any ${\epsilon }_2>0$ there is a ${\delta }_2>0$ such that $\left|g\left(x\right)-f\left(x\right)-\left(K-L\right)\right|<{\epsilon }_2$ for any $0<\left|x-a\right|<{\epsilon }_2\text{.}$

In particular, we take ${\epsilon }_2=K-L>0$ in statement 2, and we find that there is ${\delta }_2>0$ such that

3. $\left|g\left(x\right)-f\left(x\right)-\left(K-L\right)\right|<K-L$ for any $0<\left|x-a\right|<{\delta }_1\text{.}$

If we take $\delta =min\left({\delta }_1,{\delta }_2\right)$, then statements 1 and 3 are true, and for any $0<\left|x-a\right|<\delta \text{,}$ we know that

$-\left(K-L\right)<g\left(x\right)-f\left(x\right)-\left(K-L\right)<K-L\text{.}$

Adding $\left(K-L\right)$ to both sides, we obtain

$0<g\left(x\right)-f\left(x\right)<2\left(K-L\right)\text{.}$

Therefore, $f\left(x\right)<g\left(x\right)$ for $0<\left|x-a\right|<\delta \text{.}$ This inequality states that $f\left(x\right)<g\left(x\right)$ for $x\to a$, contrary to our assumption. Thus, we conclude that $K\le L\text{.}$

Q.E.D

Corollary 3.44. Squeeze Theorem.

If $g\left(x\right)\le f\left(x\right)\le h\left(x\right)$ for $x\to a$, and $\underset{x\to a}{\lim }g\left(x\right)=L=\underset{x\to a}{\lim }h\left(x\right)$, then $\underset{x\to a}{\lim }f\left(x\right)=L\text{.}$

Proof. From Theorem 3.43, we know that

$\underset{x\to a}{\lim }g\left(x\right)\le \underset{x\to a}{\lim }f\left(x\right)\le \underset{x\to a}{\lim }h\left(x\right)\text{.}$

Hence, $L\le \underset{x\to a}{\lim }f\left(x\right)\le L$, and $\underset{x\to a}{\lim }f\left(x\right)=L\text{.}$

Q.E.D

Theorem 3.45. If $\underset{x\to a}{\lim }g\left(x\right)=\infty$ and $g\left(x\right)\le f\left(x\right)$ for $x\to a$, then $\underset{x\to a}{\lim }f\left(x\right)=\infty \text{.}$

Proof. We want to show that for any $M>0$, there is a $\delta >0$ such that $f\left(x\right)>M$ for any $0<|x-a|<\delta \text{.}$

We have two assumptions:

1. there is ${\delta }_1>0$ such that $g\left(x\right)\le f\left(x\right)$ for any $0<\left|x-a\right|<{\delta }_1$
2. for $M>0$, there is a ${\delta }_2>1$ such that $g\left(x\right)>M$ for any $0<\left|x-a\right|<{\delta }_2\text{.}$

For $\delta =min\left({\delta }_1,{\delta }_2\right)$, both statements are true; and for any $0<\left|x-a\right|<\delta \text{,}$ it is the case that $f\left(x\right)\ge g\left(x\right)>M\text{.}$

Q.E.D