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# 8.2: One-Sided Limits

Difficulty Level: At Grade Created by: CK-12
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Tuscany and Sophia were out hiking. As they followed the path up the side of a hill, they discovered that there had been a washout from the recent storm. The path had been obliterated for a space of about 15 feet in front of them, whereafter it continued on up the mountain from about 10 feet higher.

How could this situation be explained with one-sided limits?

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### Guidance

Unlike the functions from prior lessons where the limit was the same from both directions, the functions we explore in this lesson may have a different limit for each "side." To evaluate these functions, we deal with each "side" separately, first evaluating what happens as the limit is approached from either the positive direction (the values of x are bigger than at the limit) or the negative direction (the values of x are smaller than at the limit) and then evaluating the other direction afterward as if it were effectively a separate function.

Sometimes the same value is approached from either "side" of the limit value, and some functions have different limits on the two sides of x = x0.

When the value of f(x) does not get closer and closer to some single value L as xx0\begin{align*}x \rightarrow x_0\end{align*} we say that the limit of the function as x approaches x0 does not exist. We may separately state the limits for each "side" of x0, but the complete function will only have a specified limit if it is the same for both sides.

For example, the two-sided limit of the complete function limx0|x|x\begin{align*}lim_{x \rightarrow 0} \frac{ |x|}{x}\end{align*} does not exist because the values of f(x) do not approach a single number as x approaches 0. We can state the one-sided limits for each side, since the values approach -1 from the left and 1 from the right.

#### Example A

Identify the limit of the function: f(x)=|x|x={1,x>01,x<0\begin{align*}f(x) = \frac{|x|} {x} = \begin{cases} 1, x > 0\\ -1, x < 0\\ \end{cases} \end{align*}

which is shown in the graph below:

Solution'

As x approaches 0 from the right, f(x) approaches 1.

On the other hand, as x approaches 0 from the left, the function f(x) approaches -1.

Since the limit is not the same from both sides, the limit of the function does not exist. However we can say that:

limx0+|x|x=1\begin{align*}\lim_{x \rightarrow 0^+} \frac{|x|} {x} = 1\end{align*}
limx0|x|x=1\begin{align*}\lim_{x \rightarrow 0^-} \frac{|x|} {x} = -1\end{align*}

Where the superscript “+” indicates a limit from the right and the superscript “-” indicates a limit from the left.

#### Example B

Consider the function f graphed in the accompanying figure and find

a) limx2f(x)\begin{align*} \lim_{x \rightarrow 2^-}f(x)\end{align*} b) limx2+f(x)\begin{align*} \lim_{x \rightarrow 2^+}f(x)\end{align*} c) limx2f(x)\begin{align*}\lim_{x \rightarrow 2}f(x)\end{align*} d) f(2)\begin{align*}f(2)\end{align*}

Solution:

a) From graph, we can see that, limx2f(x)=2\begin{align*}\lim_{x \rightarrow 2^-}f(x) = -2\end{align*}.
b) We can also see from the graph that limx2+f(x)=4\begin{align*}\lim_{x \rightarrow 2^+}f(x) = 4\end{align*}.
c) Since the limits from the right and the left are not equal (they do not approach a single value L), the limit does not exist. That is, limx2f(x)\begin{align*} \lim_{x \rightarrow 2}f(x) \end{align*} does not exist.
d) f(2)=1\begin{align*}f(2)=1\end{align*}.

#### Example C

Consider g(x)=|x2|x2|3|\begin{align*}g(x) = \frac{|x - 2|}{x - 2}|3|\end{align*} in the image below:

Find: a) limx2f(x)\begin{align*} \lim_{x \rightarrow 2^-}f(x)\end{align*} b) limx2+f(x)\begin{align*} \lim_{x \rightarrow 2^+}f(x)\end{align*} c) limx2f(x)\begin{align*}\lim_{x \rightarrow 2}f(x)\end{align*} d) f(2)\begin{align*}f(2)\end{align*}

Solution:

a) From graph, we can see that, limx2f(x)=3\begin{align*}\lim_{x \rightarrow 2^-}f(x) = -3\end{align*}.
b) We can also see from the graph that limx2+f(x)=3\begin{align*}\lim_{x \rightarrow 2^+}f(x) = 3\end{align*}.
c) Since the limits from the right and the left are not equal (they do not approach a single value L), the limit does not exist. That is, limx2f(x)\begin{align*} \lim_{x \rightarrow 2}f(x) \end{align*} does not exist.
d) f(2)=/0\begin{align*}f(2) = \not0\end{align*} undefined
Concept question wrap-up Tuscany and Sophia's path could be examined as a discontinuous function of elevation based on distance traveled along the path. For instance, if Tuscany and Sophia had traveled for 500 yards along the path before encountering the washout, then the limit of the function from the "trailhead side" would be the elevation at the edge they encountered. The function would then be undefined for the next 5 yards or so (since the path does not exist), and would pick up at 506 yards, where the elevation would be 10ft higher. If Sayber were coming down the path toward Tuscany and Sophia, from his point of view the "limit" of the elevation would be 10ft greater, and would be the lowest elevation that "his side" of the function could attain.

### Vocabulary

A one-sided-limit is a limit that only applies as the limit is approached from a specific side.

A two-sided-limit is the same value from both directions.

### Guided Practice

Questions

1) Identify the specified limit:

limx0+5x24xx\begin{align*}\lim{x\to 0^+} \frac{-5x^2 - 4x}{x}\end{align*}

2) Use the image to identify the specified limits:

a) limx2+\begin{align*}\lim_{x\to2^+}\end{align*}
b) limx2\begin{align*}\lim_{x\to2^-}\end{align*}
c) f(2)\begin{align*}f(2)\end{align*}

3) Identify the limit based on the equation:

g(x)={7;x=52;x5\begin{align*}g(x)= \begin{cases} 7; x = - 5\\ 2; x \not= -5\\ \end{cases} \end{align*}

4) Identify the limit based on the equation, use a graphing tool:

limx2+x22x+8x2=\begin{align*}\lim_{x\to2^+}\frac{-x^2 - 2x + 8}{x - 2}= \end{align*}

Solutions

1) As you can see from the graph,

a) limx+f(x)=4\begin{align*} \lim_{x \rightarrow + \infty} f(x) = 4\end{align*}, in other words, as x becomes very large, f(x) stays at the same value, 4.
b)limx+f(x)=2\begin{align*} \lim_{x \rightarrow + \infty} f(x) = -2\end{align*}, which says that as x becomes smaller and smaller, f(x) stays at the same value, -2.

2) From the image, we can see that:

a) limx2+=5\begin{align*}\lim_{x\to2^+} = 5\end{align*}
b) limx2=1\begin{align*}\lim_{x\to2^-} = -1\end{align*}
c) f(x)=2\begin{align*}f(x) = 2\end{align*} as there is a specified value for f(x)\begin{align*}f(x)\end{align*} when x=2\begin{align*}x = 2\end{align*}

3) The cases specify that if x=5\begin{align*}x = -5\end{align*} then g(x)=7\begin{align*}g(x) =7\end{align*} and if x\begin{align*}x\end{align*} is anything else, then g(x)=2\begin{align*}g(x) = 2\end{align*}

\begin{align*}\therefore\end{align*} the limit as x approaches -5 from either direction is 2.

4) To find the limit of limx3+x25x+6x3\begin{align*}\lim_{x\to3^+}\frac{x^2 - 5x + 6}{x - 3}\end{align*}

Factor the numerator: (x2)(x3)\begin{align*}(x - 2)(x - 3)\end{align*}
Now that you now have (x3)\begin{align*}(x - 3)\end{align*} in the numerator and in the denominator
Substitute 3 in for x in x2\begin{align*}x - 2\end{align*} since 3 is the number we want to evaluate
32=1\begin{align*}3 - 2 = 1\end{align*}
limx3+x25x+6x3=1\begin{align*}\therefore \lim_{x\to3^+}\frac{x^2 - 5x + 6}{x - 3} = 1\end{align*}

### Practice

Identify the limit, based on each graph.

1. limx3\begin{align*}\lim_{x\to-3^-}\end{align*}
2. limx2+\begin{align*}\lim_{x\to2^+}\end{align*}
3. limx1+\begin{align*}\lim_{x\to-1^+}\end{align*} and limx1\begin{align*}\lim_{x\to-1^-}\end{align*}
4. limx1\begin{align*}\lim_{x\to-1}\end{align*}
5. limx2\begin{align*}\lim_{x\to-2^-}\end{align*} and limx5+\begin{align*}\lim_{x\to5^+}\end{align*}

Identify the limit based on the equation:

1. limx2+x22x+8x2=\begin{align*}\lim_{x\to2^+}\frac{-x^2 - 2x + 8}{x - 2}= \end{align*}
2. g(x)={4;x=31;x3\begin{align*}g(x)= \begin{cases} 4; x = - 3\\ 1; x \not= -3\\ \end{cases}\end{align*}
3. limx0+x2+4xx=\begin{align*}\lim_{x\to0^+}\frac{-x^2 + 4x}{x}= \end{align*}
4. \begin{align*}g(x)= \begin{cases} -5; x \not= -1\\ -1; x = -1\\ \end{cases} \end{align*}
5. \begin{align*}\lim_{x\to1^+}\frac{4x^2 - x - 3}{x - 1}= \end{align*}
6. \begin{align*}f(x)= \begin{cases} 4 ; x \geq 3\\ x + 1; x < 3\\ \end{cases} \end{align*}
7. \begin{align*}\lim_{x\to0^+}\frac{x^2 - 4x}{x}= \end{align*}
8. \begin{align*}h(x)= \begin{cases} 4x + 4 ; x \not= 2\\ 1 ; x = 2\\ \end{cases} \end{align*}
9. \begin{align*}\lim_{x\to2^-}\frac{4x^2 - 7x - 2}{x - 2}= \end{align*}
10. \begin{align*}g(x)= \begin{cases} x - 5 ; x = -2\\ 4x + 1 ; x \not= -2\\ \end{cases} \end{align*}
11. \begin{align*}g(x)= \begin{cases} -3x ; \not= 3\\ -9 ; x = 3\\ \end{cases} \end{align*}
12. \begin{align*}\lim_{x \to -5^-}\frac{-3x^2 - 13x + 10}{x + 5}= \end{align*}
13. \begin{align*}f(x)= \begin{cases} x ; x = 2\\ 3x - 3 ; x \not= 2\\ \end{cases} \end{align*}

### Vocabulary Language: English

limit

limit

A limit is the value that the output of a function approaches as the input of the function approaches a given value.
one-sided limit

one-sided limit

A one-sided limit is the value that a function approaches from either the left side or the right side.
two-sided limit

two-sided limit

A two-sided limit is the value that a function approaches from both the left side and the right side.

Nov 01, 2012

Jun 08, 2015