8.6: Applications of OneSided Limits
Early in this section, we practiced finding onesided limits. In this lesson we will be using that skill and applying it with the rule of limits that says a function must have the same limit from each side in order to have a single limit.
That process allows us to first determine if a function has as limit, and then find the limit if it exists, even if we cannot actually determine the limit directly.
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 James Sousa: Graphing Quadratics
Guidance
When we wish to find the limit of a function f(x) as it approaches a point a and we cannot evaluate f(x) at a because it is undefined at that point, we can compute the function's onesided limits in order to find the desired limit. If its onesided limits are the same, then the desired limit exists and is the value of the onesided limits. If its onesided limits are not the same, then the desired limit does not exist. This technique is used in the examples below.
Conditions For a Limit to Exist (The relationship between onesided and twosided limits)


The OneSided Limit


Example A
Find the limit f(x) as x approaches 1. That is, find \begin{align*}\lim_{x \rightarrow 1} f(x)\end{align*}
\begin{align*}f(x) = \begin{cases}3  x, & x < 1 \\ 3x  x^2, & x > 1 \end{cases}\end{align*} 

Solution
Remember that we are not concerned about finding the value of f(x) at x but rather near x. So, for x < 1 (limit from the left),
\begin{align*}\lim_{x \rightarrow 1^{}} f(x) = \lim_{x \rightarrow 1^{}} (3  x) = (3  1) = 2\end{align*} 

and for x > 1 (limit from the right),
\begin{align*}\lim_{x \rightarrow 1^+} f(x) = \lim_{x \rightarrow 1^+} (3x  x^2) = 2\end{align*} 

Now since the limit exists and is the same on both sides, it follows that
\begin{align*}\lim_{x \rightarrow 1} f(x) = 2\end{align*} 

Example B
Find \begin{align*}\lim_{x \rightarrow 2} \frac{3} {x  2}\end{align*}.
Solution
From the figure below we see that \begin{align*}f(x)=\frac{3} {x  2}\end{align*} decreases without bound as x approaches 2 from the left and \begin{align*}f(x)=\frac{3} {x  2}\end{align*} increases without bound as x approaches 2 from the right.
This means that \begin{align*}\lim_{x \rightarrow 2^} \frac{3} {x  2} = \infty\end{align*} and \begin{align*}\lim_{x \rightarrow 2^+} \frac{3} {x  2} = +\infty\end{align*}. Since f(x) is unbounded (infinite) in either directions, the limit does not exist.
Example C
For an object in free fall, such as a stone falling off a cliff, the distance y(t) (in meters) that the object falls in t seconds is given by the kinematic equation y(t) = 4.9 t^{2}. The object’s velocity after 2 seconds is given by \begin{align*}v(t) = \lim_{t \rightarrow 2} \frac{y(t)  y(2)} {t  2}\end{align*}.
What is the velocity of the object after 2 seconds?
Solution
The limit is 19.6 secs. The function can be plotted on a graphing tool, and at 1.999, the graph looks like this:
You can see the result of smaller values of t, by adjusting the t slider on the active graph here: https://www.desmos.com/drive/calculator/1ombivqkdl
Vocabulary
Onesided limits are limits of a function based on an approach from each direction individually.
Guided Practice
Questions
 1) Find \begin{align*} \lim_{x \rightarrow 2} (2)\end{align*}.
 2) Find \begin{align*} \lim_{x \rightarrow 0^+} (\pi)\end{align*}.
 3) Find \begin{align*} \lim_{x \rightarrow 2} \frac{x^2  4} {x  2}\end{align*}.
 4) Find \begin{align*} \lim_{x \rightarrow 6} \frac{x  6} {x^2  36}\end{align*}.
 5) Find \begin{align*} \lim_{x \rightarrow 5} \sqrt{x^3  2x  1}\end{align*}.
Solutions

1) If a and k are real numbers, then \begin{align*}\lim_{x\rightarrow a}k = k\end{align*}.
 \begin{align*}\therefore \lim_{x \rightarrow 2} (2) = 2\end{align*}

2) If a and k are real numbers, then \begin{align*}\lim_{x\rightarrow a}k = k\end{align*}.
 \begin{align*}\therefore \lim_{x \rightarrow 0^+} (\pi) = \pi\end{align*}

3) The limit is 4, as shown in the image below. The red line approaches from values above x = 2, and the green line from below. The line is undefined where they meet. This can be examined in greater detail at: https://www.desmos.com/drive/calculator/oowpxjxeu2

4) The limit is \begin{align*}\frac{1}{12}\end{align*}
 Interact with the graph here: https://www.desmos.com/drive/calculator/jqxhysqmwy

 or make a table:

x  5
 7
 5.5
 6.5
f(x)  1/11
 1/13
 2/23
 2/25

5) The limit is \begin{align*}\sqrt{114}\end{align*} or \begin{align*}apx 10.667\end{align*}
 interact with the graph here: https://www.desmos.com/drive/calculator/2ohonznchx
Practice
Based on the graph determine if a limit exists:
Determine if a limit exists:
 \begin{align*}\lim_{x \to 0} \frac{4x^2}{x}\end{align*}
 \begin{align*} <br/> g(x)= \begin{cases} <br/> 2 ; x =  2\\ <br/> 3x + 3 ; x \not= 2\\ <br/> \end{cases} <br/> \end{align*}
 \begin{align*}\lim_{x \to 3} \frac{x^2 + 9}{x  3}\end{align*}
 \begin{align*} <br/> g(x)= \begin{cases} <br/> 3 ; x \geq 1\\ <br/> x + 4 ; x < 1\\ <br/> \end{cases} <br/> \end{align*}
 \begin{align*}\lim_{x \to 3} \frac{4x^2  15x + 9}{x  3}\end{align*}
 \begin{align*} <br/> h(x)= \begin{cases} <br/> 2; x \geq 1\\ <br/> 5x + 2 ; x < 1\\ <br/> \end{cases} <br/> \end{align*}
 \begin{align*}\lim_{x \to {1}} \frac{4x^2  4}{x + 1}\end{align*}
 \begin{align*} <br/> g(x)= \begin{cases} <br/> 3 ; x > 0\\ <br/> x  3 ; x \leq 0\\ <br/> \end{cases} <br/> \end{align*}
 \begin{align*}\lim_{x \to 4} \frac{x^2 + 5x + 4}{x + 4}\end{align*}
 \begin{align*} <br/> g(x)= \begin{cases} <br/> 3x  4 ; x = 3\\ <br/> 2x  1 ; x \not= 3\\ <br/> \end{cases} <br/> \end{align*}
 \begin{align*}\lim_{x \to 4} \frac{3x^2  15x  12}{x + 4}\end{align*}
 \begin{align*} <br/> g(x)= \begin{cases} <br/> 4 ; x \leq 3\\ <br/> 3 ; x > 3\\ <br/> \end{cases} <br/> \end{align*}
 \begin{align*}\lim_{x \to 2} \frac{2x^2  4x}{x  2}\end{align*}
 \begin{align*} <br/> f(x)= \begin{cases} <br/> 3 ; x = 1\\ <br/> 2 ; x \not= 1\\ <br/> \end{cases} <br/> \end{align*}
 \begin{align*} \lim_{x \rightarrow 3^+} \frac{3} {x  3}\end{align*}.
 Show that \begin{align*}\lim_{x \rightarrow 0^+} \left (\frac{1} {x}  \frac{1} {x^2}\right ) = \infty\end{align*}.
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