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## Section1.2Algebraic Limits

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

• 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 $x = a\text{?}$

• How do we understand limits for asymptotes and limits to infinity?

What if we only have the function and not the graph? If $x=a$ is in the domain of $f(x)$ then first simply try plugging in $a\text{,}$ this will work as long as $f(x)$ is continuous at $x=a$ (see Section 1.4 for more details). We will consider examples of functions for when $x=a$ is not in the domain.

### SubsectionLimits from a Function

###### Example1.7

Consider the functions

1. $f(x) =\displaystyle \frac{4-x^2}{x+2}\text{;}$
2. $g(x) =\displaystyle \frac{x^2-9}{x-3}\text{;}$
For both $f(x)$ and $g(x)$ consider the limits for values in and out of the domain. Note first that the domain of $f(x)$ is $x\neq-2$ and the domain of $g(x)$ is $x\neq 3\text{.}$ Thus we can evaluate limits by just plugging in values for any other point.

\begin{equation*} \lim_{x \to 0} f(x) = \frac{4}{2}=2, \end{equation*}
\begin{equation*} \lim_{x \to 2} f(x) = \frac{0}{4}=0, \end{equation*}
\begin{equation*} \lim_{x \to 0}g(x)=\frac{-9}{-3}=3, \end{equation*}
\begin{equation*} \text{and} \ \lim_{x \to 1} g(x) = \frac{-8}{-2}=4\text{.} \end{equation*}

Now let us consider what happens at $x=-2$ for $f(x)\text{.}$ Note that when you try to simply plug in $x=-2$ you get $\displaystyle \frac{0}{0}$ which is undefined. For cases of $\displaystyle \frac{0}{0}\text{,}$ first try to factor both the numerator and denominator. If both factor, cancel the common factor then try plugging in $x = a$ again.

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

What happens at $x=3$ for $g(x)\text{.}$ Again when plug in $x=3$ you get $\displaystyle \frac{0}{0}$ which is undefined.

\begin{equation*} \lim_{x \to 3} \frac{x^2-9}{x-3}=\lim_{x\to3} \frac{(x+3)(x-3)}{x-3}=\lim_{x\to3} (x+3)=6\text{.} \end{equation*}

###### Example1.8

Determine the exact value of the limit by using algebra to simplify the function.

1. $\displaystyle \lim_{x \to 1} \frac{x^2 - 1}{x-1}$
2. $\displaystyle \lim_{x \to 0} \frac{(2+x)^3 - 8}{x}$
3. $\displaystyle \lim_{x \to 0} \frac{\sqrt{x+1} - 1}{x}$
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 $\displaystyle \lim_{x \to c} f(x) = L\text{.}$ We also saw that there are cases where a function can fail to have a limit. The graphs that follow are two such examples. Figure1.9Functions $f$ and $g$ 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 Figure1.9 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 Figure1.9, 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 Figure1.9, 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 $\displaystyle \lim_{x \to 1} g(x)$ 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 Figure1.10 fails to have a limit at only two values: at $x = -2$ where the left- and right-hand limits are 2 and $-1\text{,}$ respectively and at $x = 2\text{,}$ where $\displaystyle \lim_{x \to 2^+} f(x)$ does not exist. Notice: that even at values such as $c = -1$ and $c = 3$ where there are holes in the graph, the limit still exists. Figure1.10A function $f$ demonstrates different limit behaviors.

What if you are not given the graph of $f(x)\text{?}$

###### Example1.11

Consider the piecewise function

\begin{equation*} f(x) = \begin{cases}5x^2+1\amp \text{ for $x \lt 0$ } \\ 3x+1 \amp \text{ for $x \gt 0$ } \end{cases} \end{equation*}

Let us consider $\displaystyle \lim_{x\to 0}f(x)\text{.}$ To do so we will consider the limit from either side and see if they are equal.

\begin{equation*} \lim_{x \to 0^-} f(x) = \lim_{x\to 0^-} 5x^2+1=1 \end{equation*}
\begin{equation*} \lim_{x \to 0^+} f(x) =\lim_{x \to 0^+} 3x+1=1\text{.} \end{equation*}

Since the two sides are equal then

\begin{equation*} \lim_{x \to 0} f(x) =1\text{.} \end{equation*}
###### Example1.12

Consider the piecewise function

\begin{equation*} g(x) = \begin{cases}5x-2\amp \text{ for $x \lt 2$ } \\ 4x-5 \amp \text{ for $x \gt 2$ } \end{cases} \end{equation*}

Let us consider $\displaystyle \lim_{x\to 2}g(x)\text{.}$ To do so we will consider the limit from either side and see if they are equal.

\begin{equation*} \lim_{x \to 2^-} g(x) = \lim_{x\to 2^-} 5x-2=8, \end{equation*}
\begin{equation*} \lim_{x \to 2^+} g(x) =\lim_{x \to 2^+} 4x-5=3\text{.} \end{equation*}

Since the two sides are not equal then

\begin{equation*} \lim_{x \to 2} g(x) =DNE\text{.} \end{equation*}
###### Example1.13

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 $\displaystyle \lim_{x \to c^-} f(x)$ and $\displaystyle \lim_{x \to c^+} f(x)\text{.}$

3. For each of the values $c = -2, -1, 0, 1, 2\text{,}$ determine $\displaystyle \lim_{x \to c} f(x)\text{.}$ 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. ### SubsectionSummary

• To algebraically compute the limit $\displaystyle \lim_{x \to c} f(x)\text{,}$ first try plugging in $x = c$ (if $c$ is in the domain and $f(x)$ is continuous). In the case that we have $\frac{0}{0}\text{,}$ try factoring and cancelling common factors.
• For a function $f$ defined on an interval around a number $c\text{,}$
\begin{equation*} \lim_{x \rightarrow c^-} f(x)=L_1 \end{equation*}
means that the value of $f(x)$ gets as close as we want to a number $L_1$ whenever $x$ is sufficiently close to $c$ with $x \lt c\text{,}$ assuming the value $L_1$ exists.
• Similarly, for a function $f$ defined on an interval around a number $c\text{,}$
\begin{equation*} \lim_{x \rightarrow c^+} f(x)=L_2 \end{equation*}
means that the value of $f(x)$ gets as close as we want to a number $L_2$ whenever $x$ is sufficiently close to $c$ with $x \gt c\text{,}$ assuming the value $L_2$ exists.
• The one-sided limits help to determine if a limit exists as $x$ approaches a value $c\text{.}$ More specifically, $\displaystyle \lim_{x \rightarrow c} f(x)=L$ if and only if $\displaystyle \lim_{x \rightarrow c^-} f(x)=L=\lim_{x \rightarrow c^+} f(x)$