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8.6: Applications of One-Sided Limits

Difficulty Level: At Grade Created by: CK-12

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.

Watch This

Embedded Video:

- 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 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 x 0 and must have the same value. Mathematically,
\lim_{x \rightarrow x_0} f(x) = L if and only if \lim_{x \rightarrow x_0^-} f(x) = L and \lim_{x \rightarrow x_0^+} f(x) = L .
The One-Sided Limit
If f ( x ) approaches L as x approaches x 0 from the left and from the right, then we write
\lim_{x \rightarrow x_0^+} f(x) = L
\lim_{x \rightarrow x_0^-} f(x) = L
which reads: “the limit of f ( x ) as x approaches x_0^+ (or x_0^- ) from the right (or left) is L .

Example A

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

f(x) = \begin{cases}3 - x,  & x < 1 \\3x - x^2, & x > 1 \end{cases}

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),

\lim_{x \rightarrow 1^{-}} f(x) = \lim_{x \rightarrow 1^{-}} (3 - x) = (3 - 1) = 2

and for x > 1 (limit from the right),

\lim_{x \rightarrow 1^+} f(x) = \lim_{x \rightarrow 1^+} (3x - x^2) = 2

Now since the limit exists and is the same on both sides, it follows that

\lim_{x \rightarrow 1} f(x) = 2

Example B

Find \lim_{x \rightarrow 2} \frac{3} {x - 2} .

Solution

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

This means that \lim_{x \rightarrow 2^-} \frac{3} {x - 2} = -\infty and \lim_{x \rightarrow 2^+} \frac{3} {x - 2} = +\infty . 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 v(t) = \lim_{t \rightarrow 2} \frac{y(t) - y(2)} {t - 2} .

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  \lim_{x \rightarrow -2} (2) .
2) Find  \lim_{x \rightarrow 0^+} (\pi) .
3) Find  \lim_{x \rightarrow 2} \frac{x^2 - 4} {x - 2} .
4) Find  \lim_{x \rightarrow 6} \frac{x - 6} {x^2 - 36} .
5) Find  \lim_{x \rightarrow 5} \sqrt{x^3 - 2x - 1} .

Solutions

1) If a and k are real numbers, then \lim_{x\rightarrow a}k = k .
\therefore  \lim_{x \rightarrow -2} (2) = 2
2) If a and k are real numbers, then \lim_{x\rightarrow a}k = k .
\therefore \lim_{x \rightarrow 0^+} (\pi) = \pi
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 \frac{1}{12}
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 \sqrt{114} or apx 10.667
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. \lim_{x \to 0} \frac{4x^2}{x}
  2.  <br /> g(x)= \begin{cases} <br /> -2 ; x = - 2\\ <br /> -3x + 3 ; x \not= -2\\ <br /> \end{cases} <br />
  3. \lim_{x \to 3} \frac{-x^2 + 9}{x - 3}
  4.  <br /> g(x)= \begin{cases} <br /> 3 ; x \geq -1\\ <br /> x + 4 ; x < -1\\ <br /> \end{cases} <br />
  5. \lim_{x \to 3} \frac{4x^2 - 15x + 9}{x - 3}
  6.  <br /> h(x)= \begin{cases} <br /> -2; x \geq -1\\ <br /> -5x + 2 ; x < -1\\ <br /> \end{cases} <br />
  7. \lim_{x \to {-1}} \frac{4x^2 - 4}{x + 1}
  8.  <br /> g(x)= \begin{cases} <br /> -3 ; x > 0\\ <br /> x - 3 ; x \leq 0\\ <br /> \end{cases} <br />
  9. \lim_{x \to -4} \frac{x^2 + 5x + 4}{x + 4}
  10.  <br /> g(x)= \begin{cases} <br /> -3x - 4 ; x = 3\\ <br /> -2x - 1 ; x \not= 3\\ <br /> \end{cases} <br />
  11. \lim_{x \to -4} \frac{-3x^2 - 15x - 12}{x + 4}
  12.  <br /> g(x)= \begin{cases} <br /> 4 ; x \leq -3\\ <br /> 3 ; x > -3\\ <br /> \end{cases} <br />
  13. \lim_{x \to 2} \frac{2x^2 - 4x}{x - 2}
  14.  <br /> f(x)= \begin{cases} <br /> -3 ; x = -1\\ <br /> -2 ; x \not= -1\\ <br /> \end{cases} <br />
  15.  \lim_{x \rightarrow 3^+} \frac{3} {x - 3} .
  16. Show that \lim_{x \rightarrow 0^+} \left (\frac{1} {x} - \frac{1} {x^2}\right ) = -\infty .

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Difficulty Level:

At Grade

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Date Created:

Nov 01, 2012

Last Modified:

Dec 08, 2014

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