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 onesided limits?
Watch This
Embedded Video:
 James Sousa: One Sided Limits
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 = x_{0}.
When the value of f(x) does not get closer and closer to some single value L as \begin{align*}x \rightarrow x_0\end{align*}
For example, the twosided limit of the complete function \begin{align*}lim_{x \rightarrow 0} \frac{ x}{x}\end{align*}
Example A
Identify the limit of the function: \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:


\begin{align*}\lim_{x \rightarrow 0^+} \frac{x} {x} = 1\end{align*}
limx→0+xx=1

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


\begin{align*}\lim_{x \rightarrow 0^} \frac{x} {x} = 1\end{align*}
limx→0−xx=−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) \begin{align*} \lim_{x \rightarrow 2^}f(x)\end{align*}
Solution:

a) From graph, we can see that, \begin{align*}\lim_{x \rightarrow 2^}f(x) = 2\end{align*}
limx→2−f(x)=−2 . 
b) We can also see from the graph that \begin{align*}\lim_{x \rightarrow 2^+}f(x) = 4\end{align*}
limx→2+f(x)=4 . 
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, \begin{align*} \lim_{x \rightarrow 2}f(x) \end{align*}
limx→2f(x) does not exist. 
d) \begin{align*}f(2)=1\end{align*}
f(2)=1 .
Example C
Consider \begin{align*}g(x) = \frac{x  2}{x  2}3\end{align*}
Find: a) \begin{align*} \lim_{x \rightarrow 2^}f(x)\end{align*}
Solution:

a) From graph, we can see that, \begin{align*}\lim_{x \rightarrow 2^}f(x) = 3\end{align*}
limx→2−f(x)=−3 . 
b) We can also see from the graph that \begin{align*}\lim_{x \rightarrow 2^+}f(x) = 3\end{align*}
limx→2+f(x)=3 . 
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, \begin{align*} \lim_{x \rightarrow 2}f(x) \end{align*}
limx→2f(x) does not exist.  d) \begin{align*}f(2) = \not0\end{align*} undefined
Concept question wrapup 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 onesidedlimit is a limit that only applies as the limit is approached from a specific side.
A twosidedlimit is the same value from both directions.
Guided Practice
Questions
1) Identify the specified limit:
 \begin{align*}\lim{x\to 0^+} \frac{5x^2  4x}{x}\end{align*}
2) Use the image to identify the specified limits:
 a) \begin{align*}\lim_{x\to2^+}\end{align*}
 b) \begin{align*}\lim_{x\to2^}\end{align*}
 c) \begin{align*}f(2)\end{align*}
3) Identify the limit based on the equation:
 \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:
\begin{align*}\lim_{x\to2^+}\frac{x^2  2x + 8}{x  2}= \end{align*}
Solutions
1) As you can see from the graph,
 a) \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)\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) \begin{align*}\lim_{x\to2^+} = 5\end{align*}
 b) \begin{align*}\lim_{x\to2^} = 1\end{align*}
 c) \begin{align*}f(x) = 2\end{align*} as there is a specified value for \begin{align*}f(x)\end{align*} when \begin{align*}x = 2\end{align*}
3) The cases specify that if \begin{align*}x = 5\end{align*} then \begin{align*}g(x) =7\end{align*} and if \begin{align*}x\end{align*} is anything else, then \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 \begin{align*}\lim_{x\to3^+}\frac{x^2  5x + 6}{x  3}\end{align*}
 Factor the numerator: \begin{align*}(x  2)(x  3)\end{align*}
 Now that you now have \begin{align*}(x  3)\end{align*} in the numerator and in the denominator
 Substitute 3 in for x in \begin{align*}x  2\end{align*} since 3 is the number we want to evaluate
 \begin{align*}3  2 = 1\end{align*}
 \begin{align*}\therefore \lim_{x\to3^+}\frac{x^2  5x + 6}{x  3} = 1\end{align*}
Practice
Identify the limit, based on each graph.
 \begin{align*}\lim_{x\to3^}\end{align*}
 \begin{align*}\lim_{x\to2^+}\end{align*}
 \begin{align*}\lim_{x\to1^+}\end{align*} and \begin{align*}\lim_{x\to1^}\end{align*}
 \begin{align*}\lim_{x\to1}\end{align*}
 \begin{align*}\lim_{x\to2^}\end{align*} and \begin{align*}\lim_{x\to5^+}\end{align*}
Identify the limit based on the equation:
 \begin{align*}\lim_{x\to2^+}\frac{x^2  2x + 8}{x  2}= \end{align*}
 \begin{align*}g(x)= \begin{cases} 4; x =  3\\ 1; x \not= 3\\ \end{cases}\end{align*}
 \begin{align*}\lim_{x\to0^+}\frac{x^2 + 4x}{x}= \end{align*}
 \begin{align*}g(x)= \begin{cases} 5; x \not= 1\\ 1; x = 1\\ \end{cases} \end{align*}
 \begin{align*}\lim_{x\to1^+}\frac{4x^2  x  3}{x  1}= \end{align*}
 \begin{align*}f(x)= \begin{cases} 4 ; x \geq 3\\ x + 1; x < 3\\ \end{cases} \end{align*}
 \begin{align*}\lim_{x\to0^+}\frac{x^2  4x}{x}= \end{align*}
 \begin{align*}h(x)= \begin{cases} 4x + 4 ; x \not= 2\\ 1 ; x = 2\\ \end{cases} \end{align*}
 \begin{align*}\lim_{x\to2^}\frac{4x^2  7x  2}{x  2}= \end{align*}
 \begin{align*}g(x)= \begin{cases} x  5 ; x = 2\\ 4x + 1 ; x \not= 2\\ \end{cases} \end{align*}
 \begin{align*}g(x)= \begin{cases} 3x ; \not= 3\\ 9 ; x = 3\\ \end{cases} \end{align*}
 \begin{align*}\lim_{x \to 5^}\frac{3x^2  13x + 10}{x + 5}= \end{align*}
 \begin{align*}f(x)= \begin{cases} x ; x = 2\\ 3x  3 ; x \not= 2\\ \end{cases} \end{align*}