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Applications of One-Sided Limits

Determine if a limit exists and, if so, its value.

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Practice Applications of One-Sided Limits
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Applications of One-Sided Limits

Early in this section, we practiced finding one-sided 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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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 one-sided limits in order to find the desired limit. If its one-sided limits are the same, then the desired limit exists and is the value of the one-sided limits. If its one-sided 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 one-sided and two-sided limits)
In order for the limit L of a function to exist, both of the one-sided limits must exist at x0 and must have the samevalue. Mathematically,
limxx0f(x)=L\begin{align*}\lim_{x \rightarrow x_0} f(x) = L\end{align*} if and only if limxx0f(x)=L\begin{align*}\lim_{x \rightarrow x_0^-} f(x) = L\end{align*} and limxx+0f(x)=L\begin{align*}\lim_{x \rightarrow x_0^+} f(x) = L\end{align*}.
The One-Sided Limit
If f(x) approaches L as x approaches x0from the left and from the right, then we write
limxx+0f(x)=L\begin{align*}\lim_{x \rightarrow x_0^+} f(x) = L\end{align*}
limxx0f(x)=L\begin{align*}\lim_{x \rightarrow x_0^-} f(x) = L\end{align*}
which reads: “the limit of f(x) as x approaches x+0\begin{align*}x_0^+\end{align*} (or x0\begin{align*}x_0^-\end{align*}) from the right (or left) is L.

Example A

Find the limit f(x) as x approaches 1. That is, find limx1f(x)\begin{align*}\lim_{x \rightarrow 1} f(x)\end{align*} if

f(x)={3x,3xx2,x<1x>1\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),

limx1f(x)=limx1(3x)=(31)=2\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),

limx1+f(x)=limx1+(3xx2)=2\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

limx1f(x)=2\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 t2. 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

One-sided 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:

1. \begin{align*}\lim_{x \to 0} \frac{4x^2}{x}\end{align*}
2. \begin{align*} g(x)= \begin{cases} -2 ; x = - 2\\ -3x + 3 ; x \not= -2\\ \end{cases} \end{align*}
3. \begin{align*}\lim_{x \to 3} \frac{-x^2 + 9}{x - 3}\end{align*}
4. \begin{align*} g(x)= \begin{cases} 3 ; x \geq -1\\ x + 4 ; x < -1\\ \end{cases} \end{align*}
5. \begin{align*}\lim_{x \to 3} \frac{4x^2 - 15x + 9}{x - 3}\end{align*}
6. \begin{align*} h(x)= \begin{cases} -2; x \geq -1\\ -5x + 2 ; x < -1\\ \end{cases} \end{align*}
7. \begin{align*}\lim_{x \to {-1}} \frac{4x^2 - 4}{x + 1}\end{align*}
8. \begin{align*} g(x)= \begin{cases} -3 ; x > 0\\ x - 3 ; x \leq 0\\ \end{cases} \end{align*}
9. \begin{align*}\lim_{x \to -4} \frac{x^2 + 5x + 4}{x + 4}\end{align*}
10. \begin{align*} g(x)= \begin{cases} -3x - 4 ; x = 3\\ -2x - 1 ; x \not= 3\\ \end{cases} \end{align*}
11. \begin{align*}\lim_{x \to -4} \frac{-3x^2 - 15x - 12}{x + 4}\end{align*}
12. \begin{align*} g(x)= \begin{cases} 4 ; x \leq -3\\ 3 ; x > -3\\ \end{cases} \end{align*}
13. \begin{align*}\lim_{x \to 2} \frac{2x^2 - 4x}{x - 2}\end{align*}
14. \begin{align*} f(x)= \begin{cases} -3 ; x = -1\\ -2 ; x \not= -1\\ \end{cases} \end{align*}
15. \begin{align*} \lim_{x \rightarrow 3^+} \frac{3} {x - 3}\end{align*}.
16. Show that \begin{align*}\lim_{x \rightarrow 0^+} \left (\frac{1} {x} - \frac{1} {x^2}\right ) = -\infty\end{align*}.

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