Mathematics 265 Introduction to Calculus I

Study Guide :: Unit 3

Limits

Continuous Functions

We can evaluate (find) limits by analysing the graphs of a function, but graphs are not always available or easy to obtain. So, instead, we must use the properties of the function, one of which is continuity.

In very general terms, we say that a function is continuous on an interval if the graph of the function on this interval does not break. From our basic graphs, we see that the graph of the function $f\left(x\right)=\sqrt{x}$ does not break on $\left[0,\infty \right)$, so this function is continuous on this interval. Similarly, the graphs of the trigonometric functions tell us that tangent is continuous on the interval $\left(-\pi ∕2,\pi ∕2\right)$, and many others; cosecant is continuous on $\left(0,\pi \right)$; and sine and cosine are continuous everywhere; that is, on $ℝ=\left(-\infty ,\infty \right)\text{.}$ The limit of a continuous function is easy to evaluate.

Figure 3.13. Function $f$, continuous at $a$

Figure 3.13, above, shows the graph of a function $f$ that is continuous at $a\text{.}$ We have

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

hence,

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

It could not be easier: once we know that the function is continuous at $a$, its limit at $a$ is $f\left(a\right)\text{.}$ Thus, to evaluate limits, it helps to know which functions are continuous, and where.

All the function graphs in Figures 3.2 to 3.6 break in different ways.

In Figures 3.2 to 3.4, we see that the graphs break because

$\underset{x\to a^{+}}{\lim }f\left(x\right)\ne \underset{x\to a^{-}}{\lim }f\left(x\right)\text{.}$

In Figures 3.5 and 3.6, $\underset{x\to a^{+}}{\lim }f\left(x\right)=\underset{x\to a^{-}}{\lim }f\left(x\right)$, but the graph breaks because

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

So we can see that for a function to be continuous at $a$, the following three conditions must be met:

1. $\underset{x\to a}{\lim }f\left(x\right)$ exist, that is $\underset{x\to a^{+}}{\lim }f\left(x\right)=\underset{x\to a^{-}}{\lim }f\left(x\right)\text{.}$
2. $f\left(a\right)$ is defined.
3. $\underset{x\to a}{\lim }f\left(x\right)=f\left(a\right)\text{.}$

Convince yourself that Condition 1 prevents the graph from breaking as in Figures 3.2 to 3.4, Condition 2 prevents a break as in Figure 3.5, and Condition 3 prevents a break as in Figure 3.6.

We have found the conditions for a function to be continuous at a number. We take these conditions as sufficient.

All three conditions are included in Definition 3.10, because when we say that the limit of $f$ at $a$ is equal to $f\left(a\right)$ we understand that $\underset{x\to a}{\lim }f\left(x\right)$ exist, and that $f\left(a\right)$ is defined and is equal to the limit.

When a function is continuous at every number (i.e., on $ℝ$), we say that the function is continuous everywhere.

As we mentioned earlier, sine and cosine are continuous everywhere, and from our basic graphs, we see that the identity function and the basic quadratic and cubic functions are also continuous everywhere. In fact, polynomial functions are continuous everywhere.

Example 3.19. We can apply Definition 3.10 to evaluate limits of continuous functions as follows.

1. $\underset{x\to \sqrt{3}}{\lim }6x^3+2x^2-\sqrt{6}=6{\left(\sqrt{3}\right)}^3+2{\left(\sqrt{3}\right)}^2-\sqrt{6}=18\sqrt{3}+6-\sqrt{6}\text{.}$

   This is so because ${\displaystyle P\left(x\right)=6x^3+2x^2-\sqrt{6}}$ is a polynomial function, and polynomials are continuous everywhere. In particular for our question, it is continuous at $\sqrt{3}$; hence, the limit is equal to $P\left(\sqrt{3}\right)\text{.}$

2. $\underset{x\to \frac{5\pi }{3}}{\lim }\sin x=\sin \left({\displaystyle \frac{5\pi }{3}}\right)=-{\displaystyle \frac{\sqrt{3}}{2}}\text{.}$

   Again, this is so because sine is continuous everywhere, so by Definition 3.10 the limit is equal to the value of sine at $5\pi ∕3\text{.}$

Rational functions are continuous on their domain. Hence, to evaluate the limit of a rational function at a number $a$, we need to know if the number $a$ is in the domain of the function, and then apply Definition 3.10.

Example 3.20. Let us evaluate the limit

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

We see that the function

${\displaystyle g\left(x\right)=\frac{4x^2-3}{4x+6}}$

is rational, and that the number $3$ is in its domain. [Why?] Hence, by Definition 3.10, the limit is equal to $g\left(3\right)$; that is,

${\displaystyle \underset{x\to 3}{\lim }\frac{4x^2-3}{4x+6}=\frac{4{\left(3\right)}^2-3}{4\left(3\right)+6}=\frac{11}{6}\text{.}}$

Example 3.21. The limit

${\displaystyle \underset{x\to 3}{\lim }\frac{4x^2-9}{x^2-9}}$

is also the limit of a rational function

${\displaystyle R\left(x\right)=\frac{4x^2-9}{x^2-9}\text{,}}$

but in this case, $3$ is not in the domain of this function. Hence, we cannot apply Definition 3.10. We will learn later how such limits are evaluated.

Examples 3.19 to 3.21 show that, in order to apply Definition 3.10, we must know where the functions are continuous. The answer is given by the theorem below.

Theorem 3.11. The types of functions listed below are continuous at every number of their domains.

1. polynomials
2. rational functions
3. root functions
4. trigonometric functions

In the last section of this unit, we attempt to explain why this theorem is true. For now, we simply apply this theorem, together with Definition 3.10, to evaluate limits.

Example 3.22. Let us evaluate the limit ${\displaystyle \underset{x\to \frac{5\pi }{4}}{\lim }\tan x\text{.}}$

First, we identify the trigonometric function $\tan x\text{.}$ We know from the graph of this function that $5\pi ∕4$ is in the domain of $\tan x$; hence, $\tan x$ is continuous at $5\pi ∕4$ by Theorem 3.11, and, by Definition 3.10, the limit is equal to $tan\left(5\pi ∕4\right)\text{.}$ That is,

${\displaystyle \underset{x\to \frac{5\pi }{4}}{\lim }\tan x=\tan \left(\frac{5\pi }{4}\right)=1\text{.}}$

Example 3.23. The function in the limit $\underset{x\to -5}{\lim }\sqrt[3]{x}$ is a root function, and $-5$ is in its domain (i.e., $x^{1∕3}$ is defined for any number $x$, positive or negative).

By Theorem 3.11 and Definition 3.10, the limit is equal to ${\left(-5\right)}^{1∕3}\text{.}$ That is,

$\underset{x\to -5}{\lim }\sqrt[3]{x}=\sqrt[3]{-5}={\left(-5\right)}^{1∕3}\text{.}$

Example 3.24. ${\displaystyle \underset{x\to \frac{3\pi }{4}}{\lim }\csc x=\csc \left(\frac{3\pi }{4}\right)=\sqrt{2}}$

This statement is true because $\csc x$ is continuous in its domain by Theorem 3.11, and $3\pi ∕4$ is in its domain. Thus by Definition 3.10, the limit is equal to $csc\left(3\pi ∕4\right)\text{.}$

Example 3.25. The domain of the root function in the limit $\underset{x\to -5}{\lim }\sqrt[4]{x}$ is $\left[0,\infty \right)\text{.}$ [Why?] So, the limit is undefined, since the function is not defined around $-5\text{.}$

The next two theorems extend the types of functions that are continuous on their domains.

Theorem 3.12. If  $f$ and $g$ are continuous at $a$, and $c$ is a constant, then all the functions listed below are also continuous at $a\text{.}$

1. $f+g$
2. $f-g$
3. $cf$
4. $fg$
5. ${\displaystyle \frac{f}{g}}$ if $g\left(a\right)\ne 0$

Functions $a,$ $b,$ $d$ and $e$ refer to the algebra of functions, number $c$ is the constant product of a function, which is defined as $cf\left(x\right)=c\left(f\left(x\right)\right)\text{.}$ For example, if $f\left(x\right)=\sin x+6x^2$, then $4f\left(x\right)=4\left(\sin x+6x^2\right)=4\sin x+24x^2\text{.}$

For the composition of functions, we have the following theorem.

Theorem 3.13. If the function $g$ is continuous at $a$, and the function $f$ is continuous at $g\left(a\right)$, then $f\circ g$ is continuous at $a\text{.}$

Let us see how we use Theorems 3.11 to 3.13 to decide if a function is continuous on its domain.

Example 3.26. Explain why each of the functions listed below is continuous on its domain.

1. $F(x)=\sqrt{x^2+1}$
2. $G(x)=\sqrt{x^2+1}+\sin (x^2)$
3. $H(x)=\sqrt{x^2+1}+\sin (x^2)+{\displaystyle \frac{x^3+6}{\sqrt{x^2+1}}}$

Solutions

1. The function $F$ is the composition of the functions $g\left(x\right)=x^2+1$ and $f\left(x\right)=\sqrt{x}$, as follows: $F\left(x\right)=\left(f\circ g\right)\left(x\right)\text{.}$ The function $g$ is a polynomial; hence, it is continuous everywhere. The function $f$ is continuous on $\left[0,\infty \right)$, and $x^2+1\ge 0$ for any $x$, so $f$ is continuous at $g\left(x\right)$ for any $x\text{.}$ By Theorem 3.13, above, $F$ is continuous for all $x$; that is, everywhere.

2. By Theorem 3.12, above, the function $G$ is continuous everywhere, because it is the sum of two functions, both of which are continuous everywhere; that is, $F$ (as in (b), above, and $\sin \left(x^2\right)\text{.}$ Indeed the function $\sin \left(x^2\right)$ is continuous everywhere by Theorem 3.13, because it is the composition of $\sin t$ and $x^2$, both of which are continuous everywhere by Theorem 3.12.

3. The function ${\displaystyle g\left(x\right)=\frac{x^3+6}{\sqrt{x^2+1}}}$ is the quotient of a polynomial and the function $F$, both continuous everywhere, and $F\left(x\right)\ge 0$ for all $x\text{.}$ So, by Theorem 3.12, it is true that $g$ is continuous everywhere, and the function $H$ is the sum of two everywhere continuous functions. [Which ones?] Therefore, by Theorem 3.12, it is continuous everywhere.

If you think it through carefully, you will see that the theorem below summarizes Definition 3.10 and Theorems 3.11 to 3.13. This theorem is the key to evaluating limits for continuous functions.

Theorem 3.14. If $f$ is any of the following functions

1. an algebraic function,
2. a trigonometric function,
3. the sum, difference, product or quotient of algebraic and trigonometric functions, or
4. a composition of any of the three previous types of functions,

then the function is continuous in its domain and

$\underset{x\to a}{\lim }f\left(x\right)=f\left(a\right)\text{for every }a\text{ in the domain of }f$

Example 3.27. By Theorem 3.14, all of the following functions are continuous on their domains. To find the domain, look for the variables for which the function is not defined, and eliminate them from the set of real numbers; what is left is the domain of the function. Verify that the domains listed below are correct.

Example 3.28. Use Theorem 3.14 to determine whether each of the functions below is continuous at the given number.

1. ${\displaystyle G(x)=\sec \left(2x-\pi \right)+\frac{3}{x+\pi }}$ ${\displaystyle x=\frac{\pi }{2}}$

2. $h(t)={\displaystyle \frac{6t\sin (t-1)}{t^2-1}}$ ${\displaystyle t=1}$

3. $u(s)=\csc (2s)-\sqrt{5s+1}$ $s=0$

Solutions

First observe that all of these functions are of the type listed in Theorem 3.14.

1. The function $G$ is defined at $\pi ∕2$; hence, $\pi ∕2$ is in the domain of $G$, and by Theorem 3.14, $G$ is continuous at $\pi ∕2\text{.}$
2. The function $h$ is not defined at $1$ ( $t^2-1=0$ for $t=1$); hence, $t=1$ is not in the domain of $h$, and the function is not continuous at $1\text{.}$
3. The function $u$ is not defined at $0$ because $csc\left(0\right)$ is not defined, so $0$ is not in the domain of $u$, and by Theorem 3.14, $u$ is not continuous at $0\text{.}$

Example 3.29. Finally, we will use Theorem 3.14 to evaluate the limits given below.

1. ${\displaystyle \underset{t\to -3}{\lim }\frac{t^2{\left(6-\cos t\right)}^2}{3t-\sqrt{1-t}}=-\frac{9{\left(6-\cos 3\right)}^2}{11}\text{,}}$

   because the function

   ${\displaystyle g\left(t\right)=\frac{t^2{\left(6-\cos t\right)}^2}{3t-\sqrt{1-t}}}$

   is defined at $-3$; hence, $-3$ is in its domain, and by Theorem 3.14, the function is continuous at $-3$ and the limit is equal to $g\left(-3\right)\text{.}$

2. We leave the explanation of the second result to you as an exercise:

   ${\displaystyle \underset{x\to \frac{\pi }{2}}{\lim }sec\left(2x-\pi \right)+\frac{3}{x+\pi }=sec\left(0\right)+\frac{3}{\left(3∕2\right)\pi }=1+\frac{2}{\pi }}$

Pay attention to Theorem 3.14. Most of the functions we study in this course are of the type listed in the theorem, so to evaluate a limit

$\underset{x\to a}{\lim }f\left(x\right)\text{,}$

the first thing we do is to see if we can apply Theorem 3.14. That is, we ask, “Is $f$ continuous at $a$?” or, in other words, “Is $f\left(a\right)$ defined?” If the answer is “Yes,” then we conclude that

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

Remember, however, when solving problems, that you must be careful in applying Theorem 3.14.

We also have the concept of continuity on the right and left. The graph of a function that is continuous on the right at $a$ looks like that shown in Figure 3.14, below, and the graph of a function that is continuous on the left of $a$ looks like that shown in Figure 3.15.

Figure 3.14. Function $f$, continuous on the right at $a$

Figure 3.15. Function $f$, continuous on the left at $a$

If we have understood the definition of continuity, we can see that the following definitions correspond to the graphs in Figures 3.14 and 3.15.

From Definition 3.15, we understand that when a function is continuous on the right at $a$, then $f\left(a\right)$ is defined at $a$, and $\underset{x\to a^{+}}{\lim }f\left(x\right)=f\left(a\right)\text{.}$ Similarly, when a function is continuous on the left at $a$, then $f\left(a\right)$ is defined, and $\underset{x\to a^{-}}{\lim }f\left(x\right)=f\left(a\right)\text{.}$

Example 3.30. Look carefully at the graph of Exercise 3 on page 68 of the textbook. The graph shows that this function is continuous on the left at $-2$, and on the right at $2$ and $4\text{.}$ It is not continuous either to the left or the right at $-4\text{.}$

Example 3.31. Look carefully at the graph of Exercise 4 on page 68 of the textbook. The graph shows that this function is continuous on the right at $-4$ and $2\text{.}$ It is not continuous either to the left or the right at $-2$, $4$, $6$ or $8\text{.}$

Example 3.32. Consider the following function:

${\displaystyle g\left(x\right)=\{\begin{array}{ll}1-x^2\hfill & \text{if }x\le 3\hfill \\ 3-x\hfill & \text{if }3<x<6\hfill \\ \frac{1}{x+5}\hfill & \text{if }6<x\hfill \end{array}}$

The polynomial function $1-x^2$ is continuous everywhere; in particular for our question, in the interval $\left(-\infty ,3\right)\text{.}$

Similarly $3-x$ is continuous on $\left(3,6\right)\text{.}$

${\displaystyle \frac{1}{x+5}}$ is continuous at all numbers but $-5$; hence, it is continuous on $\left(6,\infty \right)\text{.}$

Therefore, $g$ is continuous on the intervals $\left(-\infty ,3\right)$, $\left(3,6\right)$, and $\left(6,\infty \right)\text{.}$

We also see that $g\left(3\right)=1-{\left(3\right)}^2=-8$, and $g\left(6\right)$ is undefined. Hence, $g$ is not continuous at $6$, and it may be continuous on the right or left at $3\text{.}$ To see if this is the case, we must evaluate the limits on the left and right of $3\text{.}$

Observe that $x>3$ if $x\to 3^{+}$; hence, for this $x$, we have $g\left(x\right)=3-x\text{.}$ Thus,

${\displaystyle \underset{x\to 3^{+}}{\lim }g(x)=\underset{x\to 3^{+}}{\lim }3-x=3-3=0}$   (by Theorem 3.14).

If $x\to 3^{-}$, then $x<3$, and for this $x$, we have $g\left(x\right)=1-x^2\text{.}$ Thus,

${\displaystyle \underset{x\to 3^{-}}{\lim }g(x)=\underset{x\to 3^{-}}{\lim }1-x^2=-8}$   (by Theorem 3.14).

Since

${\displaystyle \underset{x\to 3^{+}}{\lim }g(x)=0\ne g(3)=\underset{x\to 3^{-}}{\lim }g(x)\text{,}}$

we conclude that the function is continuous on the left at $3$, and that it is not continuous at $3\text{.}$

Example 3.33. In this example, we will see that the piecewise function given below is continuous at $0$, discontinuous at $1$, and continuous on the right at $1\text{.}$

$h\left(x\right)=\{\begin{array}{lll}{x}^3-x^2+\sqrt{6}\hfill & \text{if}\hfill & x\le 0\hfill \\ \sqrt{6+x}\hfill & \text{if}\hfill & 0\le x<1\hfill \\ x+3\hfill & \text{if}\hfill & 1\le x\hfill \end{array}$

First, we see that $h$ is defined at $0$ and $1$, as follows: $h\left(0\right)=\sqrt{6}$ and $h\left(1\right)=4\text{.}$

$\underset{x\to 0^{+}}{\lim }h\left(x\right)=\underset{x\to 0^{+}}{\lim }\sqrt{x+6}$

and

$\underset{x\to 0^{-}}{\lim }h\left(x\right)=\underset{x\to 0^{-}}{\lim }{x}^3-x^2+\sqrt{6}=\sqrt{6}\text{.}$

Therefore, $h$ is continuous at $0$, since $\underset{x\to 0}{\lim }h\left(x\right)=\sqrt{6}=h\left(0\right)\text{.}$

$\underset{x\to 1^{-}}{\lim }h\left(x\right)=\underset{x\to 1^{-}}{\lim }\sqrt{x+6}=\sqrt{7}\text{and}\underset{x\to 1^{+}}{\lim }h\left(x\right)=\underset{x\to 1^{+}}{\lim }x+3=4$

Hence, the function is not continuous at $1$, but is continuous on the right at $1\text{.}$ [Why?]