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## Section1.2Introduction to Limits

###### Motivating Questions
• What is the mathematical notion of limit and what role do limits play in the study of functions?

• What is the meaning of the notation $\lim_{x \to a} f(x) = L\text{?}$

• How do we go about determining the value of the limit of a function at a point?

• What is a left-hand limit at $x = a$ and a right-hand limit at $x = a\text{?}$

• What does it mean graphically to say that $f$ has limit $L$ as $x \to a\text{?}$ How is this connected to having a left-hand limit at $x = a$ and having a right-hand limit at $x = a\text{?}$

• How are limits connected to continuity?

Limits are a mathematical construct we can use to describe the behavior of a function near a point. Consider the function $g$ given by the graph in Figure1.17 below. We can evaluate the function at a variety of points. For example, $g(-2)=0\text{,}$ $g(-1)=3\text{,}$ and $g(0)=1\text{.}$ Figure1.17Graph of $y = g(x)\text{.}$

A careful look at the graph above shows that $g$ has a removable discontinuity at $x=0\text{,}$ making $g(0)=1$ even though the overall shape of the graph might lead us to expect $g(0)$ to be $4\text{.}$ In fact, you would probably agree that as $x$ gets closer and closer (but NOT equal) to $0\text{,}$ $g(x)$ gets as close as we want to $4\text{.}$ This is the basic idea of a limit and we will expand on this idea in the following section.

### SubsectionThe Notion of Limit

Limits give us a way to identify a trend in the values of a function as its input variable approaches a particular value of interest. We need a precise understanding of what it means to say a function $f$ has limit $L$ as $x$ approaches $a\text{.}$

In Figure1.17, we saw that as $x$ gets closer and closer (but NOT equal) to 0, $g(x)$ gets as close as we want to the value 4. At first, this may feel counterintuitive, because the value of $g(0)$ is $1\text{,}$ not $4\text{.}$ Limits describe the behavior of a function arbitrarily close to a fixed input and are not affected by the value of the function at the fixed input. More formally,1What follows here is not what mathematicians consider the formal definition of a limit. To be completely precise, it is necessary to quantify both what it means to say as close to $L$ as we like and sufficiently close to $a\text{.}$ That can be accomplished through what is traditionally called the epsilon-delta definition of limits. That being said, the definition presented here is sufficient for the purposes of this text. we say the following.

###### Limit of a Function

If a function $f$ is defined on an interval around $c\text{,}$ except perhaps at the point $x = c\text{,}$ we define the limit of the function $f(x)$ as $x$ approaches $c$ to be a number $L$ (if one exists) such that $f(x)$ is as close to $L$ as we want whenever $x$ is sufficiently close to $c$ (but $x \neq c$). If $L$ exists, we write

\begin{equation*} \lim_{x \rightarrow c} f(x)=L\text{.} \end{equation*}

On the other hand, if as $x$ approaches $c$ we cannot make $f(x)$ as close to a single value as we would like, then we say that $f$ does not have a limit as $x$ approaches $c$.

###### Example1.18

Recall the function $g$ from the section introduction, whose graph is reproduced below. Figure1.19Graph of $y = g(x)$ For the function $g$ pictured in Figure1.19, we make the following observations:

\begin{equation*} \lim_{x \to -1} g(x) = 3, \ \lim_{x \to 0} g(x) = 4, \ \text{and} \ \lim_{x \to 2} g(x) = 1\text{.} \end{equation*}

When finding a limit from a graph, it suffices to ask if the function approaches a single value from each side of the fixed input. The function value at the fixed input is irrelevant. This reasoning explains the values of the three limits stated above.

We further observe that $g$ does not have a limit as $x$ approaches $1$ because there is a jump in the graph at $x = 1\text{.}$ If we approach $x = 1$ from the left, the function values tend to get close to 3, but if we approach $x = 1$ from the right, the function values get close to 2. There is no single number that all of these function values approach. This is why the limit of $g$ does not exist at $x = 1\text{.}$

For any function $f\text{,}$ there are typically three ways to answer the question does $f$ have a limit at $x = a\text{,}$ and if so, what is the limit? The first is to reason graphically. If we have a formula for $f(x)\text{,}$ there are two additional possibilities:

1. Evaluate the function at a sequence of inputs that approach $a$ on either side (typically using some sort of computing technology), and ask if the sequence of outputs seems to approach a single value.
2. Use the algebraic form of the function to understand the trend in its output values as the input values approach $a\text{.}$

The first approach produces only an approximation of the value of the limit, while the latter can often be used to determine the limit exactly.

###### Example1.20

For each of the following functions, we'd like to know whether or not the function has a limit at the stated $c$-values. Use both numerical and algebraic approaches to investigate and, if possible, estimate or determine the value of the limit. Compare the results with a careful graph of the function on an interval containing the points of interest.

1. $f(x) = \frac{4-x^2}{x+2}\text{;}$ $c = -1\text{,}$ $c = -2$
2. $g(x) = \sin\left(\frac{\pi}{x}\right)\text{;}$ $c = 3\text{,}$ $c = 0$
Solution

a. We first construct a graph of $f$ along with tables of values near $c = -1$ and $c = -2\text{.}$

 $x$ $f(x)$ $-0.9$ $2.9$ $-0.99$ $2.99$ $-0.999$ $2.999$ $-0.9999$ $2.9999$ $-1.1$ $3.1$ $-1.01$ $3.01$ $-1.001$ $3.001$ $-1.0001$ $3.0001$
 $x$ $f(x)$ $-1.9$ $3.9$ $-1.99$ $3.99$ $-1.999$ $3.999$ $-1.9999$ $3.9999$ $-2.1$ $4.1$ $-2.01$ $4.01$ $-2.001$ $4.001$ $-2.0001$ $4.0001$ From Table1.21, it appears that we can make $f$ as close as we want to 3 by taking $x$ sufficiently close to $-1\text{,}$ which suggests that $\lim_{x \to -1} f(x) = 3\text{.}$ This is also consistent with the graph of $y=f(x)$ seen in Figure1.23. To see this a bit more rigorously and from an algebraic point of view, consider the formula for $f\text{:}$ $f(x) = \frac{4-x^2}{x+2}\text{.}$ As $x$ approaches $-1\text{,}$ the numerator, $4-x^2\text{,}$ of $f$ approaches $4 - (-1)^2 = 3\text{,}$ and the denominator, $x+2\text{,}$ of $f$ approaches $-1 + 2 = 1\text{.}$ Hence $\lim_{x \to -1} f(x) = \frac{3}{1} = 3\text{.}$

The situation is more complicated when $x$ approaches $-2$ because $f(-2)$ is not defined. If we try to use a similar algebraic argument regarding the numerator and denominator, we observe that as $x$ approaches $-2\text{,}$ the numerator $(4-x^2)$ approaches $(4 - (-2)^2) = 0\text{,}$ and the denominator $(x+2)$ approaches $(-2 + 2) = 0\text{,}$ so as $x$ approaches $-2\text{,}$ the numerator and denominator of $f$ both tend to 0. We call $\frac{0}{0}$ an indeterminate form. This tells us that there is somehow more work to do. From Table1.22 and Figure1.23, it appears that $f$ should have a limit of $4$ at $x = -2\text{.}$

To see algebraically why this is the case, observe that

\begin{align*} \lim_{x \to -2} f(x) = \amp \lim_{x \to -2} \frac{4-x^2}{x+2}\\ = \amp \lim_{x \to -2} \frac{(2-x)(2+x)}{x+2}\text{.} \end{align*}

It is important to observe that because we are taking the limit as $x$ approaches $-2\text{,}$ we are considering $x$ values that are close, but not equal, to $-2\text{.}$ Since we never actually allow $x$ to equal $-2\text{,}$ the quotient $\frac{2+x}{x+2}$ has value 1 for every possible value of $x\text{.}$ Thus, we can simplify the most recent expression above, and find that

\begin{equation*} \lim_{x \to -2} f(x) = \lim_{x \to -2} (2-x)\text{.} \end{equation*}

This limit is now easy to determine, and its value is $4\text{.}$ Thus, from several points of view we've seen that $\lim_{x \to -2} f(x) = 4\text{.}$

b. Next we turn to the function $g\text{,}$ and construct two tables and a graph.

 $x$ $g(x)$ $2.9$ $0.84864$ $2.99$ $0.86428$ $2.999$ $0.86585$ $2.9999$ $0.86601$ $3.1$ $0.88351$ $3.01$ $0.86777$ $3.001$ $0.86620$ $3.0001$ $0.86604$
 $x$ $g(x)$ $-0.1$ $0$ $-0.01$ $0$ $-0.001$ $0$ $-0.0001$ $0$ $0.1$ $0$ $0.01$ $0$ $0.001$ $0$ $0.0001$ $0$ First, as $x$ approaches $3\text{,}$ it appears from the values in Table1.24 that the function is approaching a number between $0.86601$ and $0.86604\text{.}$ From the graph in Figure1.26 it appears that $g(x)$ approaches $g(3)$ as $x$ approaches $3\text{.}$ The exact value of $g(3) = \sin(\frac{\pi}{3})$ is $\frac{\sqrt{3}}{2}\text{,}$ which is approximately 0.8660254038. This is convincing evidence that

\begin{equation*} \lim_{x \to 3} g(x) = \frac{\sqrt{3}}{2}\text{.} \end{equation*}

As $x$ approaches $0\text{,}$ we observe that $\frac{\pi}{x}$ does not behave in an elementary way. When $x$ is positive and approaching zero, we are dividing by smaller and smaller positive values, and $\frac{\pi}{x}$ increases without bound. When $x$ is negative and approaching zero, $\frac{\pi}{x}$ decreases without bound. In this sense, as we get close to $x = 0\text{,}$ the inputs to the sine function are growing rapidly, and this leads to increasingly rapid oscillations in the graph of $g$ between $1$ and $-1\text{.}$ If we plot the function $g(x) = \sin\left(\frac{\pi}{x}\right)$ with a graphing utility and then zoom in on $x = 0\text{,}$ we see that the function never settles down to a single value near the origin, which suggests that $g$ does not have a limit at $x = 0\text{.}$

How do we reconcile the graph in Figure1.26 with Table1.25, which seems to suggest that the limit of $g$ as $x$ approaches $0$ may in fact be $0\text{?}$ The data misleads us because of the special nature of the sequence of input values $\{0.1, 0.01, 0.001, \ldots\}\text{.}$ When we evaluate $g(10^{-k})\text{,}$ we get $g(10^{-k}) = \sin\left(\frac{\pi}{10^{-k}}\right) = \sin(10^k \pi) = 0$ for each positive integer value of $k\text{.}$ But if we take a different sequence of values approaching zero, say $\{0.3, 0.03, 0.003, \ldots\}\text{,}$ then we find that

\begin{equation*} g(3 \cdot 10^{-k}) = \sin\left(\frac{\pi}{3 \cdot 10^{-k}}\right) = \sin\left(\frac{10^k \pi}{3}\right) = \frac{\sqrt{3}}{2} \approx 0.866025\text{.} \end{equation*}

That sequence of function values suggests that the value of the limit is $\frac{\sqrt{3}}{2}\text{.}$ Clearly the function cannot have two different values for the limit, so $g$ has no limit as $x$ approaches $0\text{.}$

An important lesson to take from Example1.20 is that tables can be misleading when determining the value of a limit. While a table of values is useful for investigating the possible value of a limit, we should also use other tools to confirm the value.

###### Example1.27

Estimate the value of each of the following limits by constructing appropriate tables of values. Then determine the exact value of the limit by using algebra to simplify the function. Finally, plot each function on an appropriate interval to check your result visually.

1. $\lim_{x \to 1} \frac{x^2 - 1}{x-1}$
2. $\lim_{x \to 0} \frac{(2+x)^3 - 8}{x}$
3. $\lim_{x \to 0} \frac{\sqrt{x+1} - 1}{x}$
Hint
1. $(x^2 - 1)$ can be factored.
2. Expand the expression $(2+x)^3\text{,}$ and then combine like terms in the numerator.
3. Try multiplying the given function by this fancy form of 1: $\frac{\sqrt{x+1} + 1}{\sqrt{x+1} + 1}\text{.}$
1. $2\text{.}$
2. $12\text{.}$
3. $\frac{1}{2}\text{.}$
Solution

Estimating the values of the limits with tables is straightforward and should suggest the exact values stated below.

1. \begin{align*} \lim_{x \to 1} \frac{x^2 - 1}{x-1} =\amp \lim_{x \to 1} \frac{(x+1)(x-1)}{x-1}\\ =\amp \lim_{x \to 1} (x+1) \\ =\amp 2\text{.} \end{align*}
2. \begin{align*} \lim_{x \to 0} \frac{(2+x)^3 - 8}{x} \amp = \lim_{x \to 0} \frac{8 + 12x + 6x^2 + x^3 - 8}{x}\\ \amp = \lim_{x \to 0} \frac{12x + 6x^2 + x^3}{x}\\ \amp = \lim_{x \to 0} (12 + 6x + x^2) \\ \amp = 12\text{.} \end{align*}
3. \begin{align*} \lim_{x \to 0} \frac{\sqrt{x+1} - 1}{x} \amp = \lim_{x \to 0} \frac{\sqrt{x+1} - 1}{x} \cdot \frac{\sqrt{x+1} + 1}{\sqrt{x+1} + 1}\\ \amp = \lim_{x \to 0} \frac{x+1-1}{x(\sqrt{x+1}+1)}\\ \amp = \lim_{x \to 0} \frac{1}{\sqrt{x+1}+1}\\ \amp = \frac{1}{2}\text{.} \end{align*}

### SubsectionHaving a Limit at a Point

We saw earlier that $f$ has limit $L$ as $x$ approaches $c$ provided that we can make the value of $f(x)$ as close to $L$ as we like by taking $x$ sufficiently close (but not equal to) $c\text{.}$ If so, we write $\lim_{x \to c} f(x) = L\text{.}$ We also saw that there are cases where a function can fail to have a limit. The graphs that follow are two such examples. Figure1.28Functions $f$ and $g$ that each fail to have a limit at $c = 1\text{.}$

Essentially there are two behaviors that a function can exhibit near a point where it fails to have a limit. In Figure1.28 above, at the left we see a function $f$ whose graph shows a jump at $c = 1\text{.}$ If we let $x$ approach 1 from the left side, the value of $f$ approaches 2, but if we let $x$ approach $1$ from the right, the value of $f$ tends to 3. Because the value of $f$ does not approach a single number as $x$ gets arbitrarily close to 1 from both sides, we know that $f$ does not have a limit at $c = 1\text{.}$

For such cases, we introduce the notion of left and right (or one-sided) limits.

###### One-Sided Limits

We say that $f$ has limit $L_1$ as $x$ approaches $c$ from the left and write

\begin{equation*} \lim_{x \to c^-} f(x) = L_1 \end{equation*}

provided that we can make the value of $f(x)$ as close to $L_1$ as we like by taking $x$ sufficiently close to $c$ while always having $x \lt c\text{.}$ We call $L_1$ the left-hand limit of $f$ as $x$ approaches $c\text{.}$

Similarly, we say $L_2$ is the right-hand limit of $f$ as $x$ approaches $c$ and write

\begin{equation*} \lim_{x \to c^+} f(x) = L_2 \end{equation*}

provided that we can make the value of $f(x)$ as close to $L_2$ as we like by taking $x$ sufficiently close to $c$ while always having $x \gt c\text{.}$

In the graph of $y=f(x)$ in Figure1.28, we see that

\begin{equation*} \lim_{x \to 1^-} f(x) = 2 \ \ \text{and} \ \lim_{x \to 1^+} f(x) = 3\text{.} \end{equation*}

Precisely because the left and right limits are not equal, the overall limit of $f$ as $x \to 1$ fails to exist.

For the function $g$ pictured at the right of Figure1.28, the function fails to have a limit at $c = 1$ for a different reason. While the function does not have a jump in its graph at $c = 1\text{,}$ it is still not the case that $g$ approaches a single value as $x$ approaches 1. In particular, due to the infinitely oscillating behavior of $g$ to the right of $c = 1\text{,}$ we say that the (right-hand) limit of $g$ as $x \to 1^+$ does not exist, and thus $\lim_{x \to 1} g(x)$ does not exist.

To summarize, if either a left- or right-hand limit fails to exist or if the left- and right-hand limits are not equal to each other, the overall limit does not exist.

###### Two-Sided Limit

A function $f$ has limit $L$ as $x \to c$ if and only if

\begin{equation*} \lim_{x \to c^-} f(x) = L = \lim_{x \to c^+} f(x)\text{.} \end{equation*}

That is, a function has a limit at $x = c$ if and only if both the left- and right-hand limits at $x = c$ exist and have the same value.

The function $f$ given below in Figure1.29 fails to have a limit at only two values: at $c = -2$ (where the left- and right-hand limits are 2 and $-1\text{,}$ respectively) and at $x = 2\text{,}$ where $\lim_{x \to 2^+} f(x)$ does not exist). Note well that even at values such as $c = -1$ and $c = 3$ where there are holes in the graph, the limit still exists. Figure1.29A function $f$ demonstrates different limit behaviors.
###### Example1.30

Consider a function that is piecewise-defined according to the formula

\begin{equation*} f(x) = \begin{cases}3(x+2)+2 \amp \text{ for $-3 \lt x \lt -2$ } \\ \frac{2}{3}(x+2)+1 \amp \text{ for $-2 \le x \lt -1$ } \\ \frac{2}{3}(x+2)+1 \amp \text{ for $-1 \lt x \lt 1$ } \\ 2 \amp \text{ for $x = 1$ } \\ 4-x \amp \text{ for $x \gt 1$ } \end{cases} \end{equation*}

Use the given formula to answer the following questions.

1. For each of the values $c = -2, -1, 0, 1, 2\text{,}$ compute $f(c)\text{.}$

2. For each of the values $c = -2, -1, 0, 1, 2\text{,}$ determine $\lim_{x \to c^-} f(x)$ and $\lim_{x \to c^+} f(x)\text{.}$

3. For each of the values $c = -2, -1, 0, 1, 2\text{,}$ determine $\lim_{x \to c} f(x)\text{.}$ If the limit fails to exist, explain why by discussing the left- and right-hand limits at the relevant $c$-value.

4. For which values of $c$ is the following statement true?

\begin{equation*} \lim_{x \to c} f(x) \ne f(c) \end{equation*}
5. Sketch an accurate, labeled graph of $y = f(x)\text{.}$ Be sure to carefully use open circles ($\circ$) and filled circles ($\bullet$) to represent key points on the graph, as dictated by the piecewise formula.

Hint
1. Find the interval in which $c$ lies and evaluate the function there.

2. Remember that for $\lim_{x \to c^-} f(x)\text{,}$ we only consider values of $x$ such that $x \lt c\text{.}$ Find the appropriate formula to use in the piecewise definition for $f$ to fit the values you are considering.

3. Use your work in (b) and compare left- and right-hand limits.

4. Use your work in (a) and (c).

5. Note that $f$ is piecewise linear.

1. $f(-2) = 1\text{;}$ $f(-1)$ is not defined; $f(0) = \frac{7}{3}\text{;}$ $f(1) = 2\text{;}$ $f(2) = 2\text{.}$

2. \begin{equation*} \lim_{x \to -2^-} f(x) = 2 \ \text{and} \lim_{x \to -2^+} f(x) = 1\text{.} \end{equation*}
\begin{equation*} \lim_{x \to -1^-} f(x) = \frac{5}{3} \ \text{and} \lim_{x \to -1^+} f(x) = \frac{5}{3}\text{.} \end{equation*}
\begin{equation*} \lim_{x \to 0^-} f(x) = \frac{7}{3} \ \text{and} \lim_{x \to 0^+} f(x) = \frac{7}{3}\text{.} \end{equation*}
\begin{equation*} \lim_{x \to 1^-} f(x) = 3 \ \text{and} \lim_{x \to 1^+} f(x) = 3\text{.} \end{equation*}
\begin{equation*} \lim_{x \to 2^-} f(x) = 2 \ \text{and} \lim_{x \to 2^+} f(x) = 2\text{.} \end{equation*}
3. $\lim_{x \to -2} f(x)$ does not exist. The values of the limits as $x \to c$ for $c = -1, 0, 1, 2$ are $\frac{5}{3}, \frac{7}{3}, 3, 2\text{.}$

4. $c = -2\text{,}$ $c = -1\text{,}$ and $c = 1\text{.}$

5. Solution
1. $f(-2) = \frac{2}{3}(-2+2) + 1 = 1\text{;}$ $f(-1)$ is not defined; $f(0) = \frac{2}{3}(0+2)+1 = \frac{7}{3}\text{;}$ $f(1) = 2$ (by the rule); $f(2) = 4-2 = 2\text{.}$

2. \begin{equation*} \lim_{x \to -2^-} f(x) = 2 \ \text{and} \lim_{x \to -2^+} f(x) = 1\text{.} \end{equation*}
\begin{equation*} \lim_{x \to -1^-} f(x) = \frac{5}{3} \ \text{and} \lim_{x \to -1^+} f(x) = \frac{5}{3}\text{.} \end{equation*}
\begin{equation*} \lim_{x \to 0^-} f(x) = \frac{7}{3} \ \text{and} \lim_{x \to 0^+} f(x) = \frac{7}{3}\text{.} \end{equation*}
\begin{equation*} \lim_{x \to 1^-} f(x) = 3 \ \text{and} \lim_{x \to 1^+} f(x) = 3\text{.} \end{equation*}
\begin{equation*} \lim_{x \to 2^-} f(x) = 2 \ \text{and} \lim_{x \to 2^+} f(x) = 2\text{.} \end{equation*}
3. $\lim_{x \to -2} f(x)$ does not exist because the left-hand limit is $2$ while the right-hand limit is $1\text{.}$ All of the other requested limits exist, as in each case the left- and right-hand limits exist and are equal. The respective values of the limits as $x \to c$ for $c = -1, 0, 1, 2$ are $\frac{5}{3}, \frac{7}{3}, 3, 2\text{.}$

4. For $c = -2\text{,}$ $c = -1\text{,}$ and $c = 1\text{,}$ $\lim_{x \to c} f(x) \ne f(c)\text{.}$ At $c = -2\text{,}$ the limit fails to exist, but $f(-2) = 1\text{.}$ At $c = -1\text{,}$ the limit is $\frac{5}{3}\text{,}$ but $f(-1)$ is not defined. At $c = 1\text{,}$ the limit is 3, but $f(1) = 2\text{.}$

5. ### SubsectionLimits and Continuity

Recall that informally we said a function is continuous if we can draw its graph without ever lifting our pencil from the page. Figure1.31Functions $f\text{,}$ $g\text{,}$ and $h$ that demonstrate subtly different behaviors at $c = 1\text{.}$

Using limits, we can formalize the idea of continuity. First consider the function in the left-most graph of Figure1.31 above. Note that $f(1)$ is not defined, which leads to the resulting hole in the graph of $f$ at $c = 1\text{.}$ We will naturally say that $f$ is not continuous at $c = 1$. For the function $g\text{,}$ we observe that while $\lim_{x \to 1} g(x) = 3\text{,}$ the value of $g(1)$ is $2\text{,}$ and thus the limit does not equal the function value. Here, too, we will say that $g$ is not continuous, even though the function is defined at $c = 1\text{.}$ Finally, the function $h$ appears to be the most well-behaved of all three, since at $c = 1$ its limit and its function value agree. That is,

\begin{equation*} \lim_{x \to 1} h(x) = 3 = h(1)\text{.} \end{equation*}

With no hole or jump in the graph of $h$ at $c = 1\text{,}$ we say that $h$ is continuous there. More formally, we make the following definition.

###### Continuous Function

A function $f$ is continuous at $x = c$ provided that

1. $f$ has a limit as $x \to c\text{,}$

2. $f$ is defined at $x = c\text{,}$ and

3. $\lim_{x \to c} f(x) = f(c)\text{.}$

Conditions (a) and (b) are technically contained implicitly in (c), but we state them explicitly to emphasize their individual importance. The definition says that a function is continuous at $x = c$ provided that its limit as $x \to c$ exists and equals its function value at $x = c\text{.}$ If a function is continuous at every point in an interval $[a,b]\text{,}$ we say the function is continuous on $[a,b]\text{.}$ If a function is continuous at every point in its domain, we simply say the function is continuous. Thus we note that continuous functions are particularly nice: to evaluate the limit of a continuous function at a point, all we need to do is evaluate the function.

For example, consider $p(x) = x^2 - 2x + 3\text{.}$ It can be proved that every polynomial is a continuous function at every real number, and thus if we would like to know $\lim_{x \to 2} p(x)\text{,}$ we simply compute

\begin{equation*} \lim_{x \to 2} (x^2 - 2x + 3) = 2^2 - 2 \cdot 2 + 3 = 3 = p(2)\text{.} \end{equation*}

This route of substituting an input value to evaluate a limit works whenever we know that the function being considered is continuous. Besides polynomial functions, all exponential functions and the sine and cosine functions are continuous at every point, as are many other familiar functions and combinations thereof.

###### Example1.32

Determine if each of the functions below is continuous at $x=2\text{.}$

1. $f(x)=\ln(3-x)\text{.}$
2. $g(x)=\frac{1}{x-2}\text{.}$
3. $h(x)=\begin{cases} x^2+1\amp\text{ if } x \neq 2,\\ 3\amp\text{ if } x=2.\\ \end{cases}$
Hint

Consider evaluating limits on each side and comparing that value to the value of the function at the point.

1. $f$ is continuous at $x=2\text{.}$
2. $g$ is not continuous at $x=2\text{.}$
3. $h$ is not continuous at $x=2\text{.}$
Solution

For each of these functions, we want to check that the limit exists at $x=2\text{,}$ the function is defined at $x=2\text{,}$ and these two values match.

1. We can examine the graph of $y=f(x)$ at $x=2$ or examine function values nearby $x=2$ on the left and right to find that $\lim_{x\to 2} f(x)=0\text{.}$ Evaluating $f(2)=\ln(3-2)=\ln(1)=0\text{.}$ Thus, $\lim_{x\to2}f(x)=f(2)\text{,}$ and $f$ is continuous at $x=2\text{.}$
2. Notice that the graph of $g$ has a vertical asymptote at $x=2\text{,}$ so $g(2)$ is undefined. Hence, $g$ is not continuous at $x=2\text{.}$
3. For values of $x$ near 2 (from the left and right), we have $h(x)$ getting close to 5. Therefore, $\lim_{x\to2}h(x)=5\text{.}$ However, $h(2)=3\text{.}$ Since $\lim_{x\to2}h(x)\neq h(2)\text{,}$ $h$ is not continuous at $x=2\text{.}$

### SubsectionProperties of Limits and Continuous Functions

There are several properties of limits and continuous functions that are useful to have in your toolbox. Specifically, limits and continuous functions behave well under typical mathematical operations. While these properties can be proven in detail, we proceed to only state the properties.

###### Properties of Limits

Assuming all the limits on the right-hand side exist:

• If $b$ is a constant, then $\lim\limits_{x \rightarrow c} (bf(x))=b\left(\lim\limits_{x \rightarrow c} f(x) \right)$
• $\lim\limits_{x \rightarrow c} \left( f(x)+g(x)\right)=\lim\limits_{x \rightarrow c} f(x)+\lim\limits_{x \rightarrow c}g(x)$
• $\lim\limits_{x \rightarrow c} \left( f(x) \cdot g(x)\right)=\lim\limits_{x \rightarrow c} f(x)\cdot\lim\limits_{x \rightarrow c}g(x)$
• $\lim\limits_{x \rightarrow c} \left(\frac{f(x)}{g(x)}\right)=\frac{\lim\limits_{x \rightarrow c} f(x)}{\lim\limits_{x \rightarrow c} g(x)}\text{,}$ provided $\lim\limits_{x \rightarrow c} g(x) \neq 0$
• For any constant $k\text{,}$ $\lim\limits_{x \rightarrow c} k=k$
• $\lim\limits_{x \rightarrow c} x=c$

###### Example1.33

We can use algebra to compute $\lim\limits_{x \rightarrow 1} x^2(x^3+2).$ Specifically,

\begin{align*} \lim\limits_{x \rightarrow 1} x^2(x^3+2)\amp=\left(\lim\limits_{x \rightarrow 1} x^2\right)\left(\lim\limits_{x \rightarrow 1} (x^3+2)\right)\\ \amp=\left(\lim\limits_{x \rightarrow 1} x^2\right)\left(\lim\limits_{x \rightarrow 1} x^3+\lim\limits_{x \rightarrow 1}2\right)\\ \amp=1(1+2) \\ \amp=3. \end{align*}

###### Continuity of Sums, Products, and Quotients of Functions

Suppose that $f$ and $g$ are continuous on an interval and that $b$ is a constant. Then, on that same interval,

• $bf(x)$ is continuous.
• $f(x) + g(x)$ is continuous.
• $f(x)g(x)$ is continuous.
• $\frac{f(x)}{g(x)}$ is continuous, provided $g(x) \neq 0$ on the interval.
###### Continuity of Composite Functions

If $f$ and $g$ are continuous, and if the composite function $f(g(x))$ is defined on an interval, then $f(g(x))$ is continuous on that interval.

### SubsectionSummary of Limits and Continuity

The concepts discussed in the last two sections will be important in later sections. The following is a short summary of these sections and an example that ties together the concepts of limits and continuity.

• For a function $f$ defined on an interval around a number $c\text{,}$
\begin{equation*} \lim_{x \rightarrow c} f(x)=L \end{equation*}
means that the value of $f(x)$ gets as close as we want to a number $L$ whenever $x$ is sufficiently close to $c\text{,}$ assuming the value $L$ exists.
• We define a limit from the left and a limit from the right in the same way as above, while adding the stipulation that $x\lt c$ for the left limit and $x\gt c$ for the right limit. That is, as we move $x$ sufficiently close to $c$ from the left on a number line ($x\lt c$), $f(x)$ gets as close to the limit value as we want. Similarly for the limit from the right.
• The one-sided limits help to determine if a limit exists as $x$ approaches a value $c\text{.}$ More specifically, $\lim_{x \rightarrow c} f(x)=L$ if and only if $\lim_{x \rightarrow c^-} f(x)=L=\lim_{x \rightarrow c^+} f(x)$
• Limits also help us determine the continuity of a function at a point $x=c\text{.}$ A function $f$ that has a limit as $x\rightarrow c\text{,}$ is defined at $x=c\text{,}$ and $\lim_{x\rightarrow c} f(x)=f(c)$ is continuous at $x=c\text{.}$
###### Example1.34

In this example, we take a closer look at a function whose graph we previously encountered in Figure1.29. For convenience, this graph is reproduced below in Figure1.35. Figure1.35The graph of $y = f(x)$ for Example1.34.
1. At which values of $c$ does $\lim_{x \to c} f(x)$ not exist?

2. At which values of $c$ is $f(c)$ not defined?

3. At which values of $c$ does $f$ have a limit, but $\lim_{x \to c} f(x) \ne f(c)\text{?}$

4. State all values of $c$ for which $f$ is not continuous at $x = c\text{.}$

5. Which condition is stronger, and hence implies the other: $f$ has a limit at $x = c$ or $f$ is continuous at $x = c\text{?}$ Explain, and hence complete the following sentence: If $f$ at $x = c\text{,}$ then $f$ at $x = c\text{,}$ where you complete the blanks with has a limit and is continuous, using each phrase once.

Hint
1. Consider the left- and right-hand limits at each value.

2. Carefully examine places on the graph where there's an open circle.

3. Are there locations on the graph where the function has a limit but there's a hole in the graph?

4. Remember that at least one of three conditions must fail: if the function lacks a limit, if the function is undefined, or if the limit exists but does not equal the function value, then $f$ is not continuous at the point.

5. Note that the definition of being continuous requires the limit to exist.

1. $c = -2\text{;}$ $c = +2\text{.}$

2. $c = 3\text{.}$

3. $c = -1\text{;}$ $c = 3\text{.}$

4. $c=-2\text{;}$ $c = 2\text{;}$ $c = 3\text{;}$ $c = -1\text{.}$

5. If $f$ is continuous at $x = c\text{,}$ then $f$ has a limit at $x = c\text{.}$

Solution
1. $\lim_{x \to c} f(x)$ does not exist at $c = -2$ since

\begin{equation*} \lim_{x \to -2^-} f(x) = 2 \ne -1 = \lim_{x \to -2^+}\text{,} \end{equation*}

and $\lim_{x \to c} f(x)$ does not exist at $c = 2$ since $\lim_{x \to 2^+} f(x)$ does not exist due to the infinitely oscillatory behavior of $f\text{.}$

2. The only point at which $f$ is not defined is at $c = 3\text{.}$

3. At $c = -1\text{,}$ note that $\lim_{x \to -1} f(x)$ exists (and appears to have value approximately $-3.25$), but $f(-1) = 1$ and thus $\lim_{x \to -1} f(x) \ne f(-1)\text{.}$ At $c = 3\text{,}$ we have $\lim_{x \to 3} f(x) = -2.5\text{,}$ but $f(3)$ is not defined so the limit exists but does not equal the function value.

4. Based on our work in (a), (b), and (c), $f$ is not continuous at $c=-2$ and $c = 2$ because $f$ does not have a limit at those points; $f$ is not continuous at $c = 3$ since $f$ is not defined there; and $f$ is not continuous at $c = -1$ because at that point its limit does not equal its function value.

5. If $f$ is continuous at $x = c\text{,}$ then $f$ has a limit at $x = c\text{,}$ since one of the defining properties of being continuous at $x = c$ is that the function has a limit at that input value. This shows that being continuous is a stronger condition than having a limit.