CC Algebraic Limits

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Section 1.2 Algebraic Limits

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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 $x = a ?$
How do we understand limits for asymptotes and limits to infinity?

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This section corresponds to 1.2 Algebraic Limits in the workbook.

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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,

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.

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Subsection 1.2.1 Limits from a Function

Example 1.2.1.

Consider the functions

$f (x) = \frac{4 - x^{2}}{x + 2};$
$g (x) = \frac{x^{2} - 9}{x - 3};$

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)

x \neq - 2

and the domain of

g (x)

x \neq 3 .

Thus we can evaluate limits by just plugging in values for any other point.

lim_{x \to 0} f (x) = \frac{4}{2} = 2,

lim_{x \to 2} f (x) = \frac{0}{4} = 0,

lim_{x \to 0} g (x) = \frac{- 9}{- 3} = 3,

and lim_{x \to 1} g (x) = \frac{- 8}{- 2} = 4 .

Now let us consider what happens at

x = - 2

for

f (x) .

Note that when you try to simply plug in

x = - 2

you get

\frac{0}{0}

which is undefined. For cases of

\frac{0}{0},

first try to factor both the numerator and denominator. If both factor, cancel the common factor then try plugging in

x = a

again.

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 .

What happens at

x = 3

for

g (x) .

Again when plug in

x = 3

you get

\frac{0}{0}

which is undefined.

lim_{x \to 3} \frac{x^{2} - 9}{x - 3} = lim_{x \to 3} \frac{(x + 3) (x - 3)}{x - 3} = lim_{x \to 3} (x + 3) = 6 .

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Example 1.2.2.

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

$lim_{x \to 1} \frac{x^{2} - 1}{x - 1}$
$lim_{x \to 0} \frac{(2 + x)^{3} - 8}{x}$
$lim_{x \to 0} \frac{\sqrt{x + 1} - 1}{x}$

Hint.

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

Answer.

$2 .$
$12 .$
$\frac{1}{2} .$

Solution.

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

$\begin{aligned} lim_{x \to 1} \frac{x^{2} - 1}{x - 1} = & lim_{x \to 1} \frac{(x + 1) (x - 1)}{x - 1} \\ = & lim_{x \to 1} (x + 1) \\ = & 2 . \end{aligned}$
$\begin{aligned} lim_{x \to 0} \frac{(2 + x)^{3} - 8}{x} & = lim_{x \to 0} \frac{8 + 12 x + 6 x^{2} + x^{3} - 8}{x} \\ = lim_{x \to 0} \frac{12 x + 6 x^{2} + x^{3}}{x} \\ = lim_{x \to 0} (12 + 6 x + x^{2}) \\ = 12 . \end{aligned}$
$\begin{aligned} lim_{x \to 0} \frac{\sqrt{x + 1} - 1}{x} & = lim_{x \to 0} \frac{\sqrt{x + 1} - 1}{x} \cdot \frac{\sqrt{x + 1} + 1}{\sqrt{x + 1} + 1} \\ = lim_{x \to 0} \frac{x + 1 - 1}{x (\sqrt{x + 1} + 1)} \\ = lim_{x \to 0} \frac{1}{\sqrt{x + 1} + 1} \\ = \frac{1}{2} . \end{aligned}$

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Subsection 1.2.2 Having a Limit at a Point

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We saw earlier that

f

has limit

L

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 .

If so, we write

lim_{x \to c} f (x) = L .

We also saw that there are cases where a function can fail to have a limit. The graphs that follow are two such examples.

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Figure 1.2.3. Functions $f$ and $g$ each fail to have a limit at $c = 1 .$

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Essentially there are two behaviors that a function can exhibit near a point where it fails to have a limit. In Figure 1.2.3 above, at the left we see a function

f

whose graph shows a jump at

c = 1 .

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 .

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

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One-Sided Limits.

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We say that

f

has limit

L_{1}

x

approaches

c

from the left and write

lim_{x \to c^{-}} f (x) = L_{1}

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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 < c .

We call

L_{1}

the left-hand limit of

f

x

approaches

c .

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Similarly, we say

L_{2}

is the right-hand limit of

f

x

approaches

c

and write

lim_{x \to c^{+}} f (x) = L_{2}

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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 > c .

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In the graph of

y = f (x)

in Figure 1.2.3, we see that

lim_{x \to 1^{-}} f (x) = 2 and lim_{x \to 1^{+}} f (x) = 3 .

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Precisely because the left and right limits are not equal, the overall limit of

f

x \to 1

fails to exist.

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For the function

g

pictured at the right of Figure 1.2.3, 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,

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,

we say that the (right-hand) limit of

g

x \to 1^{+}

does not exist, and thus

lim_{x \to 1} g (x)

does not exist.

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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.

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Two-Sided Limit.

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A function

f

has limit

L

x \to c

if and only if

lim_{x \to c^{-}} f (x) = L = lim_{x \to c^{+}} f (x) .

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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.

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The function

f

given below in Figure 1.2.4 fails to have a limit at only two values: at

x = - 2

where the left- and right-hand limits are 2 and

- 1,

respectively and at

x = 2,

where

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.

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Figure 1.2.4. A function $f$ demonstrates different limit behaviors.

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What if you are not given the graph of

f (x) ?

Example 1.2.5.

Consider the piecewise function

f (x) = {\begin{cases} 5 x^{2} + 1 & for x < 0 \\ 3 x + 1 & for x > 0 \end{cases}

Let us consider

lim_{x \to 0} f (x) .

To do so we will consider the limit from either side and see if they are equal.

lim_{x \to 0^{-}} f (x) = lim_{x \to 0^{-}} 5 x^{2} + 1 = 1

lim_{x \to 0^{+}} f (x) = lim_{x \to 0^{+}} 3 x + 1 = 1 .

Since the two sides are equal then

lim_{x \to 0} f (x) = 1 .

Example 1.2.6.

Consider the piecewise function

g (x) = {\begin{cases} 5 x - 2 & for x < 2 \\ 4 x - 5 & for x > 2 \end{cases}

Let us consider

lim_{x \to 2} g (x) .

To do so we will consider the limit from either side and see if they are equal.

lim_{x \to 2^{-}} g (x) = lim_{x \to 2^{-}} 5 x - 2 = 8,

lim_{x \to 2^{+}} g (x) = lim_{x \to 2^{+}} 4 x - 5 = 3 .

Since the two sides are not equal then

lim_{x \to 2} g (x) = D N E .

Example 1.2.7.

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

f (x) = {\begin{cases} 3 (x + 2) + 2 & for - 3 < x < - 2 \\ \frac{2}{3} (x + 2) + 1 & for - 2 \leq x < - 1 \\ \frac{2}{3} (x + 2) + 1 & for - 1 < x < 1 \\ 2 & for x = 1 \\ 4 - x & for x > 1 \end{cases}

Use the given formula to answer the following questions.

For each of the values $c = - 2, - 1, 0, 1, 2,$ compute $f (c) .$
For each of the values $c = - 2, - 1, 0, 1, 2,$ determine $lim_{x \to c^{-}} f (x)$ and $lim_{x \to c^{+}} f (x) .$
For each of the values $c = - 2, - 1, 0, 1, 2,$ determine $lim_{x \to c} f (x) .$ If the limit fails to exist, explain why by discussing the left- and right-hand limits at the relevant $c$ -value.
For which values of $c$ is the following statement true?

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

Hint.

Find the interval in which $c$ lies and evaluate the function there.
Remember that for $lim_{x \to c^{-}} f (x),$ we only consider values of $x$ such that $x < c .$ Find the appropriate formula to use in the piecewise definition for $f$ to fit the values you are considering.
Use your work in (b) and compare left- and right-hand limits.
Use your work in (a) and (c).
Note that $f$ is piecewise linear.

Answer.

$f (- 2) = 1;$ $f (- 1)$ is not defined; $f (0) = \frac{7}{3};$ $f (1) = 2;$ $f (2) = 2 .$
$lim_{x \to - 2^{-}} f (x) = 2 and lim_{x \to - 2^{+}} f (x) = 1 .$

$lim_{x \to - 1^{-}} f (x) = \frac{5}{3} and lim_{x \to - 1^{+}} f (x) = \frac{5}{3} .$

$lim_{x \to 0^{-}} f (x) = \frac{7}{3} and lim_{x \to 0^{+}} f (x) = \frac{7}{3} .$

$lim_{x \to 1^{-}} f (x) = 3 and lim_{x \to 1^{+}} f (x) = 3 .$

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

Solution.

$f (- 2) = \frac{2}{3} (- 2 + 2) + 1 = 1;$ $f (- 1)$ is not defined; $f (0) = \frac{2}{3} (0 + 2) + 1 = \frac{7}{3};$ $f (1) = 2$ (by the rule); $f (2) = 4 - 2 = 2 .$
$lim_{x \to - 2^{-}} f (x) = 2 and lim_{x \to - 2^{+}} f (x) = 1 .$

$lim_{x \to - 1^{-}} f (x) = \frac{5}{3} and lim_{x \to - 1^{+}} f (x) = \frac{5}{3} .$

$lim_{x \to 0^{-}} f (x) = \frac{7}{3} and lim_{x \to 0^{+}} f (x) = \frac{7}{3} .$

$lim_{x \to 1^{-}} f (x) = 3 and lim_{x \to 1^{+}} f (x) = 3 .$

$lim_{x \to 2^{-}} f (x) = 2 and lim_{x \to 2^{+}} f (x) = 2 .$
$lim_{x \to - 2} f (x)$ does not exist because the left-hand limit is $2$ while the right-hand limit is $1 .$ 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 $\frac{5}{3}, \frac{7}{3}, 3, 2 .$
For $c = - 2,$ $c = - 1,$ and $c = 1,$ $lim_{x \to c} f (x) \neq f (c) .$ At $c = - 2,$ the limit fails to exist, but $f (- 2) = 1 .$ At $c = - 1,$ the limit is $\frac{5}{3},$ but $f (- 1)$ is not defined. At $c = 1,$ the limit is 3, but $f (1) = 2 .$

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Subsection 1.2.3 Summary

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To algebraically compute the limit $lim_{x \to c} f (x),$ first try plugging in $x = c$ (if $c$ is in the domain and $f (x)$ is continuous). In the case that we have $\frac{0}{0},$ try factoring and cancelling common factors.
For a function $f$ defined on an interval around a number $c,$
$lim_{x \to c^{-}} f (x) = L_{1}$
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 $x < c,$ assuming the value $L_{1}$ exists.
Similarly, for a function $f$ defined on an interval around a number $c,$
$lim_{x \to c^{+}} f (x) = L_{2}$
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 $x > c,$ assuming the value $L_{2}$ exists.
The one-sided limits help to determine if a limit exists as $x$ approaches a value $c .$ More specifically, $lim_{x \to c} f (x) = L$ if and only if $lim_{x \to c^{-}} f (x) = L = lim_{x \to c^{+}} f (x)$

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Exercises 1.2.4 Exercises

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1. Limits on a piecewise graph.

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Use the figure below, which gives a graph of the function

f (x),

to give values for the indicated limits. If a limit does not exist, enter none.

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(a)

lim_{x \to - 2} f (x)

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(b)

lim_{x \to 0} f (x)

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(c)

lim_{x \to 2} f (x)

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(d)

lim_{x \to 4} f (x)

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2. Limits for a piecewise formula.

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For the function

f (x) = {\begin{cases} x^{2} - 3, & 0 \leq x < 3 \\ - 1, & x = 3 \\ 3 x - 3, & 3 < x \end{cases}

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use algebra to find each of the following limits:

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lim_{x \to 3^{+}} f (x) =

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lim_{x \to 3^{-}} f (x) =

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lim_{x \to 3} f (x) =

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(For each, enter DNE if the limit does not exist.)

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Sketch a graph of

f (x)

to confirm your answers.

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3. Calculating Limits of Rational Functions.

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Evaluate the limit

lim_{x \to 5} \frac{x - 5}{x^{2} - 5 x}

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Enter DNE if the limit does not exist.

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Limit =

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4. One-Sided Limits.

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Evaluate the limits.

f (x) = {\begin{cases} 7 x + 9 & x \leq 1 \\ x^{2} - 1 & x > 1 \end{cases}

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Enter DNE if the limit does not exist.

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lim_{x \to 1^{-}} f (x)

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lim_{x \to 1^{+}} f (x)

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lim_{x \to 1} f (x)

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5. Evaluating a limit algebraically.

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Evaluate the limit

lim_{x \to - 7} \frac{x^{2} - 49}{x + 7}

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If the limit does not exist enter DNE.

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Limit =

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