6.1 A (more) Formal Introduction to Limits

Limits from Graphs

Even though we didn't introduce the limit concept or notation until Chapter 2, we first encounter the underlying concept much earlier. Recall in Example 1.1.4 we were given the graph of a function $w=F(v)$:

The hole in the graph tells us that even though $F(1)$ is undefined, we'd expect $F(1)$ to be $-3$ based on what's happening with the graph near the point $(1,-3)$. Using limit notation, we'd write $\underset{v\to 1}{lim}F(v)=-3$. We take a moment below to better define what we mean when use the limit notation.

Some remarks about Definition 6.1 are in order. Note that the business about $f$ being defined on `an open interval containing $x=a$' is there to guarantee that we have the appropriate `room' for inputs $x$ to approach $a$ from either direction.⁴ (For now, we'll just assume we all understand what the word `approach' means in this context and let a Calculus class explain how this is more precisely codified mathematically.)

The phrase `except possibly at $x=a$' which immediately follows means the limit doesn't concern itself with what is actually happening at $x=a$. The function $f$ may or may not be defined at $x=a$. Indeed, if $\underset{x\to a}{lim}f(x)=L$, $f(a)$ could be $L$, $f(a)$ could be a number different than $L$ or $f(a)$ could not be defined.

This drives home the principle difference between the precalculus notion of `$f(a)$' and the Calculus notion of `$\underset{x\to a}{lim}f(x)$': `$\underset{x\to a}{lim}f(x)$' is what we expect $f(a)$ to be - which may or may not agree with what $f(a)$, if $f(a)$ is even defined.

For example, using the graph from Example 1.1.4, we write $\underset{v\to 0}{lim}F(v)=-4$ since as $v\to 0$, we see $w=F(v)\to -4$. In this particular case, $F(0)=-4$ so we get from $F$ at $v=0$ what we expect to get.⁵

For another example, consider the graphs of the functions $f$, $g$, and $h$ below near $x=2$. Through a precalculus lens, each of these functions is different at $x=2$: $f(2)=3$, $g(2)=1$, and $h(2)$ is undefined. Through a Calculus lens, however, all three of these functions are behaving identically as $x$ approaches $2$: $\underset{x\to 2}{lim}f(x)=3$, $\underset{x\to 2}{lim}g(x)=3$, and $\underset{x\to 2}{lim}h(x)=3$.

$$y=f(x)$$

$$y=g(x)$$

$$y=h(x)$$

Next let's head to Section 1.2 and revisit Example 1.2.1 in a piecewise-defined function is used to model matinee admission prices at a local theater:

What can be said about $\underset{A\to 6}{lim}p(A)$? Remember, $\underset{A\to 6}{lim}p(A)$ is what we would expect $p(6)$ to be by analyzing $p$ as $A\to 6$, ignoring what is happening at $A=6$. If $A<6$, $p(A)$ is always $5.75$, so, based on this information, we'd expect $p(6)$ to be $5.75$. If $A>6$, then $p(A)$ is always $7.25$, so we'd expect $p(6)$ to be $7.25$. Since Definition 6.1 requires the $p(A)$ values to approach a single value $L$ as $A\to 6$, we'd say in this case that $\underset{A\to 6}{lim}p(A)$ does not exist.

Even though $\underset{A\to 6}{lim}p(A)$ does not exist, we've used so-called `one-sided' limit notation in Chapters 3 and 4 which we can apply here. Specifically, we write $\underset{A\to 6^{-}}{lim}p(A)=5.75$ and $\underset{A\to 6^{+}}{lim}p(A)=7.25$ to more precisely record the behavior of $p$ as we approach $A=6$ from either direction.⁶

In order for the (two-sided) limit to exist, both one-sided limits need to exist, be equal, and vice-versa. This is recorded in the following theorem.

It's time for an example.

Another use of limits we've seen is to codify unbounded behavior. Since $\mathrm{\infty }$ and $-\mathrm{\infty }$ aren't real numbers, we used limit notation to help us describe end behavior (as $x\to -\mathrm{\infty }$ or $x\to \mathrm{\infty }$) and unbounded function behavior ($f(x)\to -\mathrm{\infty }$ or $f(x)\to \mathrm{\infty }$.) Let's take a moment to think about what it means to write $\underset{x\to \mathrm{\infty }}{lim}f(x)=\mathrm{\infty }$. How does one `approach' infinity anyhow?

Let's consider $\underset{x\to \mathrm{\infty }}{lim}{x}^2=\mathrm{\infty }$. What me mean here is that as $x$ grows larger and larger (without bound), $f(x)=x^2$ follows suit. To prove something like this, we'd need to show that for any `arbitrarily large' real number, $N$, we can find some threshold $M$ so that if the inputs, $x>M$, the outputs, $f(x)>N$. For example, if we set $N=10000$, then to guarantee $f(x)=x^2>10000$, we can solve and get $x>\sqrt{10000}=100$. So provided $x>100$, $f(x)>10000$. In this case, $N=10000$ and $M=\sqrt{10000}=100$. In general, if $x>\sqrt{N}$, $x^2>N$, which justifies us writing $\underset{x\to \mathrm{\infty }}{lim}{x}^2=\mathrm{\infty }$.

We can adjust the inequality signs in the sort of argument⁸ above to direct $x$ or $f(x)$ to either $\mathrm{\infty }$ or $-\mathrm{\infty }$. Doing so gives us the (formal) definitions of below.

We'll explore Definition 6.3 more in the Exercises. In the meantime, the reader is encouraged to take some time and think about the inequalities in Definition 6.3 and how they force the corresponding graphical behavior showcased below:

$$\underset{x\to \mathrm{\infty }}{lim}f(x)=\mathrm{\infty }$$

$$\underset{x\to \mathrm{\infty }}{lim}f(x)=-\mathrm{\infty }$$

$$\underset{x\to -\mathrm{\infty }}{lim}f(x)=\mathrm{\infty }$$

$$\underset{x\to -\mathrm{\infty }}{lim}f(x)=-\mathrm{\infty }$$

Combining the ideas of what it means for $x$ or $f(x)$ to approach (finite) real numbers along with our (more precise notion) of what it means for $x$ or $f(x)$ to approach $-\mathrm{\infty }$ or $\mathrm{\infty }$, we can mix and match to produce expressions and graphs containing vertical and horizontal asymptotes such as the ones depicted below:

$$\underset{x\to 5^{-}}{lim}f(x)=\mathrm{\infty }$$

$$\underset{x\to 0}{lim}f(x)=-\mathrm{\infty }$$

$$\underset{x\to \mathrm{\infty }}{lim}f(x)=3$$

We would be remiss in our duties as (pre)Calculus instructors if we failed to point out that even though we've used notation `$=\mathrm{\infty }$' in expressions like $\underset{x\to 5^{-}}{lim}f(x)=\mathrm{\infty }$ above, since $\mathrm{\infty }$ is not a real number, technically, $\underset{x\to 5^{-}}{lim}f(x)$ does not exist. The `$=\mathrm{\infty }$' here just codifies better the manner in which the limit fails to exist.

Our last example of this section turns the tables and has you construct the graph of function given information provided by limits.

Limit Properties and an Introduction to Continuity

Let $f(x)=6$. Consider $\underset{x\to 5}{lim}f(x)=\underset{x\to 5}{lim}6$. Since the function values are unchanging, there is no other value other than `$6$' to expect from $f$ so it stands to reason that $\underset{x\to 5}{lim}f(x)=\underset{x\to 5}{lim}6=6$. Indeed, for any real number $a$, $\underset{x\to a}{lim}6=6$. In general, if $f(x)=c$ is a constant function, $\underset{x\to a}{lim}f(x)=\underset{x\to a}{lim}c=c$. The formal proof of this fact requires a formal definition of limit,but for now, we'll just take it as true.

Next, let's consider $f(x)=x$. Consider $\underset{x\to 5}{lim}f(x)=\underset{x\to 5}{lim}x$. What do we expect the value of `$x$' to be as $x\to 5$? Well, `$5$'. Indeed, it can be proved that $\underset{x\to a}{lim}x=a$ for all real numbers, $a$.

What about $\underset{x\to 5}{lim}(x+6)$? Since $\underset{x\to 5}{lim}x=5$ and $\underset{x\to 5}{lim}6=6$, it stands to reason that

$$\underset{x\to 5}{lim}(x+6)=\underset{x\to 5}{lim}x+\underset{x\to 5}{lim}6=5+6=11,$$

which is indeed the case. It turns out that in most cases, limits do respect arithmetic:

For those interested, the Scalar Multiple Rule and Power Rule are grouped with the Product Rule since they both follow directly from the Product Rule. For instance, using the Product Rule,

$$\underset{x\to a}{lim}\left[cf(x)\right]=\underset{x\to a}{lim}c\underset{x\to a}{lim}f(x)=c\underset{x\to a}{lim}f(x).$$

For powers, note that ${\left[f(x)\right]}^2=f(x)f(x)$ so that

$$\underset{x\to a}{lim}{\left[f(x)\right]}^2=\underset{x\to a}{lim}\left[f(x)f(x)\right]=\underset{x\to a}{lim}f(x)\underset{x\to a}{lim}f(x)=LL=L^2.$$

Once this is established, we can use the fact that ${\left[f(x)\right]}^3=f(x){\left[f(x)\right]}^2$ and the product rule again to get

$$\underset{x\to a}{lim}{\left[f(x)\right]}^3=\underset{x\to a}{lim}\left[f(x){\left[f(x)\right]}^2\right]=\underset{x\to a}{lim}f(x)\underset{x\to a}{lim}{\left[f(x)\right]}^2=L{L}^2=L^3.$$

Continuing in this manner gives us the Power Rule.¹⁰

A note regarding the Rules for Radicals: since $\sqrt{N}$ is not real if $N<0$, we have to be careful about limits involving even-indexed radicals (or exponents which indicate even-indexed radicals.) For example, consider $\underset{x\to 5}{lim}\sqrt{5-x}$. Since this is a `two-sided' limit, we must consider both $x\to 5^{-}$ and $x\to 5^{+}$.

As $x\to 5^{-}$, the radicand, $(5-x)>0$ so $\sqrt{5-x}$ is defined as a real number. More specifically, as $x\to 5^{-}$, the quantity $(5-x)\to 0^{+}$ so $\underset{x\to 5^{-}}{lim}\sqrt{5-x}=0$. On the other hand, if $x\to 5^{+}$, the quantity $(5-x)<0$, and $\sqrt{5-x}$ is no longer a real number. Therefore, $\underset{x\to 5^{+}}{lim}\sqrt{5-x}$, and, hence, $\underset{x\to 5}{lim}\sqrt{5-x}$ does not exist.

Note the Real Number Powers rule can be thought as a generalization of the Power Rule, Quotient Rule, and Rules for Radicals for the case $L>0$. Recall that positive rational number exponents can be defined in terms of natural number powers and radicals as: $x^{\frac{m}{n}}={\left(\sqrt[n]{x}\right)}^m$. Negative exponents can be defined in terms of quotients: $x^{-\frac{m}{n}}=\frac{1}{x^{\frac{m}{n}}}$. For the Real Number Exponents rule, we are generalizing the exponents to any real number but keeping the stipulation that $L>0$ to make sure the resulting answer is defined.¹¹

We put the limit properties to good use in the following example.

It is worth noting that we could have arrived at the same (correct) answer to Example 6.1.3 by evaluating $f(3)$: $f(3)=\frac{3\sqrt{3+1}}{{(3)}^2+6-4}=\frac{6}{11}$. That's really the power of Theorem 6.2. Under `nice' circumstances,¹² Theorem 6.2 allows us to compute limits using direct substitution. Functions with this property have a familiar name.

This is not the first time we've mentioned this property of functions. Indeed, we've discussed continuity albeit in graphical terms throughout Chapters 1 through 4. In those chapters, we described continuous functions as those whose graphs are connected meaning they have `no holes or breaks' in them. It is a great exercise to compare the description given in Definition 6.4 to the graphical description to see how those two ideas mesh.

In a standard Calculus course, you'll explore properties of continuous functions more extensively. For our purposes here, polynomial, and, more generally, rational functions are continuous on their domains, as well as the functions we encountered in Chapter 4. Indeed, so long as we avoid the usual domain pitfalls, combining continuous functions via the standard four operation function arithmetic or using function composition results in a continuous function. This means in order to evaluate limits of these functions, we may use Definition 6.4 and simply evaluate the function at the corresponding value.

It is worth noting that despite each `piece' of the piecewise-defined function $f$ in Example 6.1.4 being continuous, the pieces don't match up at $x=2$ causing what is called a discontinuity. A discontinuity is a place where a function is not continuous. The particular variety of discontinuity appearing here is usually called a `jump' discontinuity - a type of discontinuity belonging to a larger class of `non-removable' or `essential' discontinuities. We'll point out other types of discontinuities as we encounter them.¹⁴

We close this section with an example that ties (most of) the fundamental concepts of limits and their calculations together.

The above reasoning is sound and is true in general. We'll be getting a lot of use out of the following:

Exercises

1. Consider the complete graph of the function $f$ below. Use the graph to find the indicated values.

   If a limit fails to exist, state that is the case or use the symbols `$\mathrm{\infty }$' or `$-\mathrm{\infty }$' appropriately.

   NOTE: The graph has a vertical asymptote $x=3$.

   

   • $\underset{x\to -3^{-}}{lim}f(x)$
   • $\underset{x\to -3^{+}}{lim}f(x)$
   • $\underset{x\to -3}{lim}f(x)$
   • $f(-3)$
   • $\underset{x\to 0}{lim}f(x)$
   • $f(0)$
   • $\underset{x\to 3^{-}}{lim}f(x)$
   • $\underset{x\to 3^{+}}{lim}f(x)$
   Answer

   • $$\underset{x\to -3^{-}}{lim}f(x)=3$$
   • $$\underset{x\to -3^{+}}{lim}f(x)=0$$
   • $\underset{x\to -3}{lim}f(x)$ d.n.e.
   • $f(-3)=0$

   • $\underset{x\to 0}{lim}f(x)=9$
   • $f(0)$
   • $$\underset{x\to 3^{-}}{lim}f(x)=0$$
   • $$\underset{x\to 3^{+}}{lim}f(x)=-\mathrm{\infty }$$
2. 1. Explain why if $\underset{x\to a}{lim}f(x)$ exist, then so do $\underset{x\to a^{-}}{lim}f(x)$ and $\underset{x\to a^{+}}{lim}f(x)$.
   2. Find an instance¹⁹ where $\underset{x\to a^{-}}{lim}f(x)$ and $\underset{x\to a^{+}}{lim}f(x)$ both exist but $\underset{x\to a}{lim}f(x)$ does not.
   Answer

   1. If $\underset{x\to a}{lim}f(x)$ exists, say $\underset{x\to a}{lim}f(x)=L$ then the $f(x)$ values approach $L$ as $x\to a$ from both directions. Hence both one-sided limits exist. In particular, $$\underset{x\to a^{-}}{lim}f(x)=L$$ and $$\underset{x\to a^{+}}{lim}f(x)=L$$.
   2. In Example 6.1.1, $$\underset{x\to -1^{-}}{lim}f(x)=0$$ and $$\underset{x\to -1^{+}}{lim}f(x)=4$$ both exist but $\underset{x\to -1}{lim}f(x)$ does not because the two one-sided limits are not equal.

3. Consider the complete graph of the function $g$ below. Use the graph to find the indicated values.

   If a limit fails to exist, state that is the case or use the symbols `$\mathrm{\infty }$' or `$-\mathrm{\infty }$' appropriately.

   NOTE: The graph has a vertical asymptote $x=-2$ and a horizontal asymptote $y=1$.

   

   • $\underset{x\to -\mathrm{\infty }}{lim}g(x)$
   • $\underset{x\to -2^{-}}{lim}g(x)$
   • $\underset{x\to -2^{+}}{lim}g(x)$
   • $\underset{x\to \mathrm{\infty }}{lim}g(x)$
   • $\underset{x\to 0}{lim}g(x)$
   • $\underset{x\to 2^{-}}{lim}g(x)$
   • $g(2)$
   • $\underset{x\to 2^{+}}{lim}g(x)$
   Answer

   • $$\underset{x\to -\mathrm{\infty }}{lim}g(x)=1$$
   • $$\underset{x\to -2^{-}}{lim}g(x)=-\mathrm{\infty }$$
   • $$\underset{x\to -2^{+}}{lim}g(x)=\mathrm{\infty }$$
   • $$\underset{x\to \mathrm{\infty }}{lim}g(x)=-\mathrm{\infty }$$

   • $\underset{x\to 0}{lim}g(x)=2$
   • $$\underset{x\to 2^{-}}{lim}g(x)=1.5$$
   • $g(2)=1.5$
   • $$\underset{x\to 2^{+}}{lim}g(x)=0$$

For Exercises -, find the limit analytically using Exercise 6.1.5 as a guide. If a limit fails to exist, state that is the case or use the symbols `$\mathrm{\infty }$' or `$-\mathrm{\infty }$' appropriately.

1. $\underset{x\to 2}{lim}\frac{2x^2+x-3}{x^2-1}$
   Answer

   $$\underset{x\to 2}{lim}\frac{2x^2+x-3}{x^2-1}=\frac{7}{3}$$

2. $\underset{x\to 1}{lim}\frac{2x^2+x-3}{x^2-1}$
   Answer

   $$\underset{x\to 1}{lim}\frac{2x^2+x-3}{x^2-1}=\frac{5}{2}$$

3. $\underset{x\to -1}{lim}\frac{2x^2+x-3}{x^2-1}$
   Answer

   $\underset{x\to -1}{lim}\frac{2x^2+x-3}{x^2-1}$ does not exist

4. $\underset{x\to -1^{+}}{lim}\frac{2x^2+x-3}{x^2-1}$
   Answer

   $$\underset{x\to -1^{+}}{lim}\frac{2x^2+x-3}{x^2-1}=\mathrm{\infty }$$

5. $\underset{x\to 3}{lim}\frac{x^2-3x}{x^2-x-6}$
   Answer

   $$\underset{x\to 3}{lim}\frac{x^2-3x}{x^2-x-6}=\frac{3}{5}$$

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

   $$\underset{x\to 3}{lim}\frac{x^2-6x}{x^2-6x+9}=-\mathrm{\infty }$$

For Exercises -, use the piecewise definition of absolute value, Definition 1.9 to help you find the limit analytically. If a limit fails to exist, state that is the case or use the symbols `$\mathrm{\infty }$' or `$-\mathrm{\infty }$' appropriately.

1. $\underset{x\to 3^{-}}{lim}\frac{|3x-x^2|}{x-3}$
   Answer

   $$\underset{x\to 3^{-}}{lim}\frac{|3x-x^2|}{x-3}=-3$$

2. $\underset{x\to 2^{+}}{lim}\frac{|6-3x|}{x^2-4x+4}$
   Answer

   $$\underset{x\to 2}{lim}\frac{|6-3x|}{x^2-4x+4}=\mathrm{\infty }$$

For Exercises -, simplify the complex fraction in order to help you find the limit analytically.²⁰ If a limit fails to exist, state that is the case or use the symbols `$\mathrm{\infty }$' or `$-\mathrm{\infty }$' appropriately.

1. $\underset{x\to 1}{lim}\frac{{\displaystyle \frac{x}{x-2}}+1}{x-1}$
   Answer

   $$\underset{x\to 1}{lim}\frac{{\displaystyle \frac{x}{x-2}}+1}{x-1}=-2$$

2. $\underset{x\to 2}{lim}\frac{{\displaystyle \frac{2x}{x+2}}-1}{x-2}$
   Answer

   $$\underset{x\to 2}{lim}\frac{{\displaystyle \frac{2x}{x+2}}-1}{x-2}=\frac{1}{4}$$

3. $\underset{h\to 0}{lim}\frac{{\displaystyle \frac{1}{2(x+h)-1}}-{\displaystyle \frac{1}{2x-1}}}{h}$
   Answer

   $$\underset{h\to 0}{lim}\frac{{\displaystyle \frac{1}{2(x+h)-1}}-{\displaystyle \frac{1}{2x-1}}}{h}=-\frac{2}{{(2x-1)}^2}$$

For Exercises -, rationalize the numerator of the fraction in order to help you find the limit analytically.²¹ If a limit fails to exist, state that is the case or use the symbols `$\mathrm{\infty }$' or `$-\mathrm{\infty }$' appropriately.

1. $\underset{x\to -1}{lim}\frac{\sqrt{x+5}-2}{x+1}$
   Answer

   $$\underset{x\to -1}{lim}\frac{\sqrt{x+5}-2}{x+1}=\frac{1}{4}$$

2. $\underset{h\to 0}{lim}\frac{\sqrt{4+h}-2}{h}$
   Answer

   $$\underset{h\to 0}{lim}\frac{\sqrt{4+h}-2}{h}=\frac{1}{4}$$

3. $\underset{h\to 0}{lim}\frac{\sqrt{2x+2h-1}-\sqrt{2x-1}}{h}$
   Answer

   $$\underset{h\to 0}{lim}\frac{\sqrt{2x+2h-1}-\sqrt{2x-1}}{h}=\frac{1}{\sqrt{2x-1}}$$

In Exercises -, find the limit analytically. Use the symbols `$\mathrm{\infty }$' and `$-\mathrm{\infty }$' as appropriate.

1. $\underset{x\to \mathrm{\infty }}{lim}\frac{3x-4}{2x+1}$
   Answer

   $$\underset{x\to \mathrm{\infty }}{lim}\frac{3x-4}{2x+1}=\frac{3}{2}$$

2. $\underset{x\to -\mathrm{\infty }}{lim}\frac{1-2x}{x-5}$
   Answer

   $$\underset{x\to -\mathrm{\infty }}{lim}\frac{1-2x}{x-5}=-2$$

3. $\underset{x\to -\mathrm{\infty }}{lim}\frac{2x-1}{x^2+4}$
   Answer

   $$\underset{x\to -\mathrm{\infty }}{lim}\frac{2x-1}{x^2+4}=0$$

4. $\underset{x\to \mathrm{\infty }}{lim}\frac{\sqrt{4x^2+x-1}}{1-x}$
   Answer

   $$\underset{x\to \mathrm{\infty }}{lim}\frac{\sqrt{4x^2+x-1}}{1-x}=-2$$

5. $\underset{x\to -\mathrm{\infty }}{lim}\frac{\sqrt{4x^2+x-1}}{1-x}$
   Answer

   $$\underset{x\to -\mathrm{\infty }}{lim}\frac{\sqrt{4x^2+x-1}}{1-x}=2$$

6. $\underset{x\to \mathrm{\infty }}{lim}\frac{x^2+2x+3}{4-x}$.
   Answer

   $$\underset{x\to \mathrm{\infty }}{lim}\frac{x^2+2x+3}{4-x}=-\mathrm{\infty }$$

7. Let $f(x)=\frac{x}{\lfloor x\rfloor }$ where `$\lfloor x\rfloor$' is the greatest integer (or floor) function.²²

   Fill in the blanks below to help you analyze $\underset{x\to 0}{lim}f(x)$.

   1. If $-1<x<0$, then $\lfloor x\rfloor =\underset{¯}{}$. So we can rewrite $f(x)=\frac{x}{\lfloor x\rfloor }=\underset{¯}{}$.
   2. Using part (a), we can find $$\underset{x\to 0^{-}}{lim}f(x)=\underset{x\to 0^{-}}{lim}\underset{¯}{}=\underset{¯}{}$$.
   3. If $0<x<1$, then $\lfloor x\rfloor =\underset{¯}{}$, hence $f(x)=\frac{x}{\lfloor x\rfloor }$ does not exist as $x\to 0^{+}$.
   4. Putting parts (b) and (c) together, we have that $\underset{x\to 0}{lim}f(x)\underset{¯}{}$
   5. Graph $f(x)=\frac{x}{\lfloor x\rfloor }$ on desmos near $x=0$ to confirm your answers.
   Answer

   1. If $-1<x<0$, then $\lfloor x\rfloor =-1$. So we can rewrite $f(x)=\frac{x}{\lfloor x\rfloor }=-x$.
   2. Using part (a), we can find $$\underset{x\to 0^{-}}{lim}f(x)=\underset{x\to 0^{-}}{lim}-x=0$$.
   3. If $0<x<1$, then $\lfloor x\rfloor =0$, hence $f(x)=\frac{x}{\lfloor x\rfloor }$ does not exist as $x\to 0^{+}$.
   4. Putting parts (b) and (c) together, we have that $\underset{x\to 0}{lim}f(x)$ does not exist.

8. Sketch the graph of a function which satisfies all of the following criteria:

   • $\underset{x\to -\mathrm{\infty }}{lim}f(x)=\mathrm{\infty }$
   • $\underset{x\to 4^{-}}{lim}f(x)=6$
   • $\underset{x\to 4^{+}}{lim}f(x)=-\mathrm{\infty }$
   • $\underset{x\to \mathrm{\infty }}{lim}f(x)=0$
   Answer

   Answers vary.

9. Sketch the graph of a function $f$ which satisfies all of the following criteria:

   • $\underset{x\to -\mathrm{\infty }}{lim}f(x)=2$
   • $\underset{x\to 0^{-}}{lim}f(x)=\mathrm{\infty }$
   • $\underset{x\to 0^{+}}{lim}f(x)=-\mathrm{\infty }$
   • $\underset{x\to 2^{-}}{lim}f(x)=3$
   • $\underset{x\to 2^{+}}{lim}f(x)=0$
   • $\underset{x\to \mathrm{\infty }}{lim}f(x)=-\mathrm{\infty }$
   Answer

   Answers vary.

10. A function is said to be continuous from the left at $x=a$ if $\underset{x\to a^{-}}{lim}f(x)=f(a)$. Likewise, a function is said to be continuous from the right at $x=a$ if $\underset{x\to a^{+}}{lim}f(x)=f(a)$.

    1. Explain why $r(x)=\sqrt{5-x}$ is not continuous at $x=5$. Is $r$ continuous from the left at $x=5$? From the right? Explain.
    2. Explain why the floor function²³ $F(x)=\lfloor x\rfloor$ is not continuous at $x=117$. Is $F$ continuous from the left at $x=117$? From the right? Explain.
    3. If a function $f$ is continuous at $x=a$, explain why $f$ is continuous from both the left and the right at $x=a$. Is the converse true? That is, if $f$ is continuous from both the left and the right at $x=a$, is $f$ continuous at $x=a$?
    4. Compare and contrast your answers in this exercise to those in Exercise.
    Answer

    1. The function $r(x)=\sqrt{5-x}$ is not continuous at $x=5$ since $r$ is undefined if $x>5$, so $\underset{x\to 5^{+}}{lim}r(x)$ does not exist. However, $r$ is continuous from the left at $x=5$ since

       $$\underset{x\to 5^{-}}{lim}r(x)=\underset{x\to 5^{-}}{lim}\sqrt{5-x}=0=\sqrt{5-5}=r(5)$$.

    2. The function $F(x)=\lfloor x\rfloor$ is not continuous at $x=117$ since $$\underset{x\to {117}^{-}}{lim}\lfloor x\rfloor =116$$ but $$\underset{x\to {117}^{+}}{lim}\lfloor x\rfloor =117$$. However, $F$ is continuous from the right at $x=117$ since $$\underset{x\to {117}^{+}}{lim}\lfloor x\rfloor =117=\lfloor 117\rfloor =F(117)$$.
    3. If $f$ is continuous at $x=a$, then $$\underset{x\to a}{lim}f(x)=f(a)$$. This means $$\underset{x\to a^{-}}{lim}f(x)=f(a)$$ and $$\underset{x\to a^{+}}{lim}f(x)=f(a)$$, so $f$ is continuous from both directions at $x=a$. The converse is also true since ff $$\underset{x\to a^{-}}{lim}f(x)=f(a)$$ and $$\underset{x\to a^{+}}{lim}f(x)=f(a)$$, then $$\underset{x\to a}{lim}f(x)=f(a)$$.
    4. The difference between the scenario here and that in Exercise is that here, we know what each of the one-sided limits are: $f(a)$. They can't be different numbers like they could be in Example 6.1.1

11. Consider the table of values below:

    $$\begin{array}{cc}& \\ \hfill x& f(x)\\ & \\ \hfill -0.001& 1.9\\ & \\ \hfill -0.0001& 1.99\\ & \\ \hfill -0.00001& 1.999\\ & \\ \hfill -0.000001& 1.9999\\ & \\ \hfill 0.000001& -10000\\ & \\ \hfill 0.00001& -1000\\ & \\ \hfill 0.0001& -100\\ & \\ \hfill 0.001& -10\end{array}$$

    It turns out that $\underset{x\to 0}{lim}f(x)=117$. How is this possible assuming the data in the table is correct?

    Answer

    We are told nothing of the function on the interval $$(-0.000001,0.000001)$$ so the function has plenty of opportunities to approach $117$.

12. In this exercise, we use Definition 6.3 to prove $\underset{x\to \mathrm{\infty }}{lim}{x}^3=\mathrm{\infty }$ and $\underset{x\to -\mathrm{\infty }}{lim}{x}^3=-\mathrm{\infty }$:

    1. Solve the following inequalities:

       • $x^3>1000$
       • $x^3>100000$
       • $x^3>{10}^{99}$
       • $x^3>N$
    2. Show that for $N>0$, if $x>\sqrt[3]{N}$, then $x^3>N$. Write a sentence (or two!) which uses your work and Definition 6.3 to prove $\underset{x\to \mathrm{\infty }}{lim}{x}^3=\mathrm{\infty }$.
    3. Repeat a similar argument to prove $\underset{x\to -\mathrm{\infty }}{lim}{x}^3=-\mathrm{\infty }$
    Answer

    1. • $x^3>1000$ for $x>10$.
       • $x^3>1000000$ for $x>100$.
       • $x^3>{10}^{99}$ for $x>{10}^{33}$.
       • $x^3>N$ for $x>\sqrt[3]{N}$.
    2. If $x>\sqrt[3]{N}$, then $$x^3>{\left(\sqrt[3]{N}\right)}^3=N$$. Per Definition 6.3, given $N>0$, choose $M=\sqrt[3]{N}$. If $x>M$, then $x^3>M^3=N$. Hence, $\underset{x\to \mathrm{\infty }}{lim}{x}^3=\mathrm{\infty }$.
    3. Given $N<0$, choose $M=\sqrt[3]{N}$. If $x<M$, then $$x^3<M^3={\left(\sqrt[3]{N}\right)}^3=N$$. Hence, $\underset{x\to -\mathrm{\infty }}{lim}{x}^3=-\mathrm{\infty }$.