1.3 Graphical Limits

$\newenvironment {prompt}{}{} \newcommand {\bart }[0]{BART} \newcommand {\ffrac }[2]{\frac {\text {\footnotesize $#1$}}{\text {\footnotesize $#2$}}} \newcommand {\npnoround }[0]{\nprounddigits {-1}} \newcommand {\npnoroundexp }[0]{\nproundexpdigits {-1}} \newcommand {\npunitcommand }[1]{\ensuremath {\mathrm {#1}}} \newcommand {\RR }[0]{\mathbb R} \newcommand {\R }[0]{\mathbb R} \newcommand {\N }[0]{\mathbb N} \newcommand {\Z }[0]{\mathbb Z} \newcommand {\d }[0]{\mathop {}\!d} \newcommand {\l }[0]{\ell } \newcommand {\ddx }[0]{\frac {d}{\d x}} \newcommand {\zeroOverZero }[0]{\ensuremath {\boldsymbol {\tfrac {0}{0}}}} \newcommand {\inftyOverInfty }[0]{\ensuremath {\boldsymbol {\tfrac {\infty }{\infty }}}} \newcommand {\zeroOverInfty }[0]{\ensuremath {\boldsymbol {\tfrac {0}{\infty }}}} \newcommand {\zeroTimesInfty }[0]{\ensuremath {\small \boldsymbol {0\cdot \infty }}} \newcommand {\inftyMinusInfty }[0]{\ensuremath {\small \boldsymbol {\infty -\infty }}} \newcommand {\oneToInfty }[0]{\ensuremath {\boldsymbol {1^\infty }}} \newcommand {\zeroToZero }[0]{\ensuremath {\boldsymbol {0^0}}} \newcommand {\inftyToZero }[0]{\ensuremath {\boldsymbol {\infty ^0}}} \newcommand {\numOverZero }[0]{\ensuremath {\boldsymbol {\tfrac {\#}{0}}}} \newcommand {\dfn }[0]{\textbf } \newcommand {\unit }[0]{\mathop {}\!\mathrm } \newcommand {\eval }[1]{ #1 \bigg |} \newcommand {\seq }[1]{\left ( #1 \right )} \newcommand {\epsilon }[0]{\varepsilon } \newcommand {\iff }[0]{\Leftrightarrow } \DeclareMathOperator {\arccot }{arccot} \DeclareMathOperator {\arcsec }{arcsec} \DeclareMathOperator {\arccsc }{arccsc} \DeclareMathOperator {\si }{Si} \DeclareMathOperator {\proj }{proj} \DeclareMathOperator {\scal }{scal} \newcommand {\tightoverset }[2]{\mathop {#2}\limits ^{\vbox to -.5ex{\kern -0.75ex\hbox {$#1$}\vss }}} \newcommand {\arrowvec }[0]{\overrightarrow } \newcommand {\vec }[0]{\mathbf } \newcommand {\veci }[0]{{\boldsymbol {\hat {\imath }}}} \newcommand {\vecj }[0]{{\boldsymbol {\hat {\jmath }}}} \newcommand {\veck }[0]{{\boldsymbol {\hat {k}}}} \newcommand {\vecl }[0]{\boldsymbol {\l }} \newcommand {\utan }[0]{\vec {\hat {t}}} \newcommand {\unormal }[0]{\vec {\hat {n}}} \newcommand {\ubinormal }[0]{\vec {\hat {b}}} \newcommand {\dotp }[0]{\bullet } \newcommand {\cross }[0]{\boldsymbol \times } \newcommand {\grad }[0]{\boldsymbol \nabla } \newcommand {\divergence }[0]{\grad \dotp } \newcommand {\curl }[0]{\grad \cross } \newcommand {\vector }[1]{\left \langle #1\right \rangle } \newcommand {\HyperFirstAtBeginDocument }[0]{\AtBeginDocument } \newcommand {\epstopdfsetup }[0]{\setkeys {ETE}} \newcommand {\GetTitleStringSetup }[0]{\setkeys {gettitlestring}} \newcommand {\Sectionformat }[2]{#1}$

In this section we use the graph of a function to find limits.

Finding Limits Graphically

Here are some detailed, lecture style videos on graphical limits:

In this section, functions will be presented graphically. Recall that the graph of a function must pass the vertical line test which states that a vertical line can intersect the graph of a function in at most one point. To understand graphical representations of functions, consider the following graph of a function, $y = f(x)$ .

The graph of a function is created by letting the $x$ -coordinate represent the input of the function and the $y$ -coordinate represent the corresponding output, i.e., $y = f(x)$ . The general form of a point on the graph of $y = f(x)$ is $(x_0, f(x_0))$ for any input value, $x_0$ , in the domain of $f$ .

Notice that the point $(-1,2)$ is on the graph of $f$ , shown above. This means that $f(-1) = 2$ . Similarly, since the points $(3,-2)$ and $(5,1)$ are also on the graph, we have $f(3) = -2$ and $f(5) = 1$ .

We now consider several examples of limits of functions presented graphically.

For the function, $f$ , whose graph is shown below, determine the value of the one-sided limit, $\lim _{x\to 3^-} f(x).$ Note that this is called a one-sided limit because $x$ is approaching $3$ from the left hand side.

Solution: Since $x \to 3^-$ we know that the $x$ -value is approaching $3$ and we also know that $x<3$ . Looking at the graph, we can see that for $x$ -values slightly less than $3$ (to the left of 3), the $y$ -values on the graph are very close to $2$ . And as the $x$ -value moves toward 3, the corresponding $y$ -value on the graph moves toward 2. We can express this using limits as

$\lim _{x\to 3^-} f(x) = 2.$

In this example, it is important to note that the open circle at the point $(3,2)$ indicates that that point is not on the graph. Furthermore, there are no points on the graph with $x$ -coordinate equal to 3, which means that $f(3)$ is undefined. Thus, we see in this example that a function can have a limit as $x$ approaches a value that is not in the domain of the function. Interactivity note: if you click on the graph, a dot will appear on the graph. You can then drag that dot along the curve with your mouse.

Use the graph of $y = f(x)$ given below to determine $\lim _{x\to -4^+} f(x).$

$x$ is to the right of -4

The y-coordinates determine the limit

Click on the graph and move the dot

The value of the limit is $\answer [given]{3}$

For the function, $f$ , whose graph is shown below, determine the value of the two-sided limit, $\lim _{x\to 3} f(x).$ Note that this is called a two-sided limit because $x$ can approach $3$ from either the left hand side or the right hand side.

Solution: Since $x \to 3$ we know that the $x$ -value is approaching $3$ and we also know that $x$ can be on either side of $3$ (but not equal to 3). Looking at the graph, we can see that for $x$ -values either slightly less than $3$ or slightly greater than $3$ , the $y$ -values on the graph are very close to $2$ . Thus,

$\lim _{x\to 3} f(x) = 2.$

In this example it is important to observe that, even though the function value at $x= 3$ is $4$ , as indicated by the dot at the point $(3,4)$ , the limit of the function as $x$ approaches $3$ is $2$ and not $4$ . The limit of a function is determined by the behavior of the function near the indicated $x$ -value and not at that $x$ -value.

Use the graph of $y = f(x)$ given below to determine $\lim _{x\to 3} f(x).$

This is a two-sided limit

The y-coordinates determine the limit

Click on the graph and move the dot

The value of the limit is $\answer [given]{1}$

For the function, $f$ , whose graph is shown below, find the one-sided limits, $\lim _{x \to 1^-}f(x), \lim _{x \to 1^+}f(x),$ and discuss the two-sided limit, $\lim _{x \to 1} f(x).$

Solution: As in example 1, we can determine the value of the left hand limit, $\lim _{x \to 1^-}f(x)$ , by observing that if $x$ is on the left side of 1 and very close to 1 then the $y$ -values on the graph are very close to -1 and hence,

$\lim _{x \to 1^-}f(x) = -1.$

For the right hand limit $\lim _{x \to 1^+}f(x)$ , we inspect the $y$ -values on the graph that correspond to $x$ -values on the right side of 1 and very close to $1$ . We see that these $y$ -values are very close to 2, hence

$\lim _{x \to 1^+}f(x) = 2.$

We see that the one-sided limits as $x$ approaches 1 have different values, and we therefore conclude that the two-sided limit, $\lim _{x \to 1}f(x)$ , does not exist. We write

$\lim _{x \to 1}f(x) \;\;\text {DNE}.$

In general, when the one-sided limits are different then the corresponding two-sided limit does not exist.

Use the graph of $y = f(x)$ given below to determine $\lim _{x\to -1^-} f(x), \lim _{x\to -1^+} f(x) \ \text {and}$ $\lim _{x\to -1} f(x).$

The y-coordinates determine the limit

Click on the graph and move the dot

It is possible that a limit DNE

The value are $\lim _{x\to -1^-} f(x)=\answer [given]{-2}$ , $\lim _{x\to -1^+} f(x)=\answer [given]{3}$ and $\lim _{x\to -1} f(x)=\answer [given]{DNE}$

For the function, $f$ , whose graph is shown below, find the one-sided limit, $\lim _{x \to 3^+}f(x).$

Solution: From the graph, we can see that as the $x$ -values approach 3 from the right hand side, the $y$ -values decrease without bound. In this case we write

$\lim _{x \to 3^+}f(x) = -\infty .$

To describe the phenomenon of an infinite limit as $x$ approaches a finite value (in this case, 3), we say that the line $x = 3$ is a vertical asymptote for the graph of $f$ .

Use the graph of $y = f(x)$ given below to determine $\lim _{x\to 2^-} f(x).$

$x$ is to the left of 2

The y-coordinates determine the limit

Click on the graph and move the dot

Type infinity for $\infty$ and -infinity for $-\infty$

The value of the limit is $\answer [given]{\infty }$

For the function, $f$ , whose graph is shown below, find the limit, $\lim _{x \to \infty }f(x).$ This is an example of a limit at infinity and it is used to help us describe the end behavior of the function.

Solution: Since $x$ is approaching infinity, we look for a pattern on the right end of the graph. From the graph, we can see that the $y$ -values are getting closer and closer to $2$ as the $x$ values increase without bound. We can conclude that

$\lim _{x \to \infty }f(x) = 2,$

and we can describe the end behavior by saying that the line $y = 2$ is a horizontal asymptote for the graph of $f$ .

Use the graph of $y = f(x)$ given below to determine $\lim _{x\to -\infty } f(x).$

Look at the left end of the graph

The y-coordinates determine the limit

Click on the graph and move the dot

The value of the limit is $\answer [given]{0}$

Here is a problem that encompasses all of the ideas in this section.

Use the above graph of $y = f(x)$ to answer the following questions.

What are the following function values (undefined is a possibility)? $f(-4) = \answer [given]{3}, \ f(-1) = \answer [given]{-1}$ $f(0) = \answer [given]{undefined}, \ f(1) = \answer [given]{2}$ $f(4) = \answer [given]{3}, \ f(6) = \answer [given]{-3}$ Find the following one-sided limits. (You can click on the graph a drag a point along it.) $\lim _{x\to -4^-} f(x) = \answer [given]{3}, \ \lim _{x\to -4^+} f(x) = \answer [given]{1}$ $\lim _{x\to -1^-} f(x) = \answer [given]{4}, \ \lim _{x\to -1^+} f(x) = \answer [given]{-1}$ Type infinity for $\infty$ and -infinity for $-\infty$ . $\lim _{x\to 0^-} f(x) = \answer [given]{-\infty }, \ \lim _{x\to 0^+} f(x) = \answer [given]{\infty }$ $\lim _{x\to 1^-} f(x) = \answer [given]{1}, \ \lim _{x\to 1^+} f(x) = \answer [given]{2}$ $\lim _{x\to 4^-} f(x) = \answer [given]{3}, \ \lim _{x\to 4^+} f(x) = \answer [given]{2}$ $\lim _{x\to 6^-} f(x) = \answer [given]{0}, \ \lim _{x\to 6^+} f(x) = \answer [given]{2}$ Find the following limits at infinity. (Click on the grid and drag it if necessary. Press the home button on the right to reset the grid.) $\lim _{x\to -\infty } f(x) = \answer [given]{2}, \ \lim _{x\to \infty } f(x) = \answer [given]{1}$