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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 a limit, and then find the limit if it exists, even if we cannot actually determine the limit directly.

### 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 limx23x2\begin{align*}\lim_{x \rightarrow 2} \frac{3} {x - 2}\end{align*}.

Solution:

From the figure below we see that f(x)=3x2\begin{align*}f(x)=\frac{3} {x - 2}\end{align*} decreases without bound as x approaches 2 from the left and f(x)=3x2\begin{align*}f(x)=\frac{3} {x - 2}\end{align*} increases without bound as x approaches 2 from the right.

This means that limx23x2=\begin{align*}\lim_{x \rightarrow 2^-} \frac{3} {x - 2} = -\infty\end{align*} and limx2+3x2=+\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 v(t)=limt2y(t)y(2)t2\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 -->

### Guided Practice

1) Find limx2(2)\begin{align*} \lim_{x \rightarrow -2} (2)\end{align*}.
2) Find limx0+(π)\begin{align*} \lim_{x \rightarrow 0^+} (\pi)\end{align*}.
3) Find limx2x24x2\begin{align*} \lim_{x \rightarrow 2} \frac{x^2 - 4} {x - 2}\end{align*}.
4) Find limx6x6x236\begin{align*} \lim_{x \rightarrow 6} \frac{x - 6} {x^2 - 36}\end{align*}.
5) Find limx5x32x1\begin{align*} \lim_{x \rightarrow 5} \sqrt{x^3 - 2x - 1}\end{align*}.

1) If a and k are real numbers, then limxak=k\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*}\approx 10.667\end{align*}.
Interact with the graph here: https://www.desmos.com/drive/calculator/2ohonznchx

### Explore More

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*}.

### Answers for Explore More Problems

To view the Explore More answers, open this PDF file and look for section 8.6.

### Vocabulary Language: English

continuity

Continuity for a point exists when the left and right sided limits match the function evaluated at that point. For a function to be continuous, the function must be continuous at every single point in an unbroken domain.

Continuous

Continuity for a point exists when the left and right sided limits match the function evaluated at that point. For a function to be continuous, the function must be continuous at every single point in an unbroken domain.

Jump discontinuities

Inverse functions are functions that 'undo' each other. Formally $f(x)$ and $g(x)$ are inverse functions if $f(g(x)) = g(f(x)) = x$.

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

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

Removable discontinuities

Removable discontinuities are also known as holes. They occur when factors can be algebraically canceled from rational functions.

Removable discontinuity

Removable discontinuities are also known as holes. They occur when factors can be algebraically canceled from rational functions.

two-sided limit

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

### Explore More

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