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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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One Sided Limits and Continuity

A one sided limit is exactly what you might expect; the limit of a function as it approaches a specific x\begin{align*}x\end{align*} value from either the right side or the left side. One sided limits help to deal with the issue of a jump discontinuity and the two sides not matching.

Is the following piecewise function continuous?

f(x)=x2x<13  x=1x24 1<x\begin{align*}f(x)=\begin{cases} -x -2 \quad x < 1 \\ \quad -3 \ \ \quad x=1 \\ x^2 -4 \quad \ 1< x \end{cases}\end{align*}

#### Watch This

http://www.youtube.com/watch?v=3iZUK15aPE0 James Sousa: Determining Limits and One-Sided Limits Graphically

#### Guidance

A one sided limit can be evaluated either from the left or from the right. Since left and right are not absolute directions, a more precise way of thinking about direction is “from the negative side” or “from the positive side”. The notation for these one sided limits is:

limxaf(x), limxa+f(x)\begin{align*}\lim \limits_{x \to a^-}f(x), \ \lim \limits_{x \to a^+}f(x)\end{align*}

The negative in the superscript of a\begin{align*}a\end{align*} is not an exponent. Instead it indicates from the negative side. Likewise the positive superscript is not an exponent, it just means from the positive side. When evaluating one sided limits, it does not matter what the function is doing at the actual point or what the function is doing on the other side of the number. Your job is to determine what the height of the function should be using only evidence on one side.

You have defined continuity in the past as the ability to draw a function completely without lifting your pencil off of the paper. You can now define a more rigorous definition of continuity.

If both of the one sided limits equal the value of the function at a given point, then the function is continuous at that point. In other words, a function is continuous at a\begin{align*}a\end{align*} if:

limxaf(x)=f(a)=limxa+f(x)\begin{align*}\lim\limits_{x \to a^-}f(x)=f(a)=\lim\limits_{x \to a^+}f(x)\end{align*}

Example A

What are the one sided limits at -5, -1, 3 and 5?

Solution:

limx5f(x)=1\begin{align*}\lim\limits_{x \to -5^-}f(x)=1\end{align*}

limx5+f(x)=2\begin{align*}\lim\limits_{x \to -5^+}f(x)=2\end{align*}

limx1f(x)=2\begin{align*}\lim\limits_{x \to -1^-}f(x)=2\end{align*}

limx1+f(x)=3\begin{align*}\lim\limits_{x \to -1^+}f(x)=3\end{align*}

limx3f(x)=3\begin{align*}\lim\limits_{x \to 3^-}f(x)=3\end{align*}

limx3+f(x)=2\begin{align*}\lim\limits_{x \to 3^+}f(x)=-2\end{align*}

limx5f(x)=2\begin{align*}\lim\limits_{x \to 5^-}f(x)=-2\end{align*}

limx5+f(x)=DNE\begin{align*}\lim\limits_{x \to -5^+}f(x)=DNE\end{align*}

Example B

Evaluate the one sided limit at 4 from the negative direction numerically.

f(x)=x27x+12x4\begin{align*}f(x)=\frac{x^2-7x+12}{x-4}\end{align*}

Solution: When creating the table, only use values that are smaller than 4.

 x\begin{align*}x\end{align*} 3.9 3.99 3.999 \begin{align*}f(x)\end{align*} 0.9 0.99 0.999

\begin{align*}\lim\limits_{x \to 4^-}\left(\frac{x^2-7x+12}{x-4}\right)=1\end{align*}

Example C

Evaluate the following limits.

1. \begin{align*}\lim\limits_{x \to 3^-}(4x-3)\end{align*}
2. \begin{align*}\lim\limits_{x \to 2^+}\left(\frac{1}{x-2}\right)\end{align*}
3. \begin{align*}\lim\limits_{x \to 1^+}\left(\frac{x^2+2x-3}{x-1}\right)\end{align*}

Solution: Most of the time one sided limits are the same as the corresponding two sided limit. The exceptions are when there are jump discontinuities, which normally only happen with piecewise functions, and infinite discontinuities, which normally only happen with rational functions.

a. \begin{align*}\lim\limits_{x \to 3^-}(4x-3)=4\cdot3-3=12-3=9\end{align*}

b. \begin{align*}\lim\limits_{x \to 2^+}\left(\frac{1}{x-2}\right)=DNE\end{align*} or \begin{align*}\infty\end{align*}

The reason why \begin{align*}\infty\end{align*} is preferable in this case is because the two sides of the limit disagree. One side goes to negative infinity and the other side goes to positive infinity (see the graph below). If you just indicate DNE then you are losing some perfectly good information about the nature of the function.

c. \begin{align*}\lim\limits_{x \to 1^+}\left(\frac{x^2+2x-3}{x-1}\right)=\lim\limits_{x \to 1^+}\left(\frac{(x-1)(x+3)}{(x-1)}\right)=\lim\limits_{x \to 1^+}(x+3)=1+3=4\end{align*}

Concept Problem Revisited

In order to confirm or deny that the function is continuous, graphical tools are not accurate enough. Sometimes jump discontinuities can be off by such a small amount that the pixels on the display of your calculator will not display a difference. Your calculator will certainly not display removable discontinuities.

\begin{align*}f(x)=\begin{cases} -x -2 \quad x < 1 \\ \quad -3 \ \ \quad x=1 \\ x^2 -4 \quad \ 1< x \end{cases}\end{align*}

You should note that on the graph, everything to the left of 1 is continuous because it is just a line. Next you should note that everything to the right of 1 is also continuous for the same reason. The only point to check is at \begin{align*}x=1\end{align*}.  To check continuity, explicitly use the definition and evaluate all three parts to see if they are equal.

• \begin{align*}\lim\limits_{x \to a^-}f(x)=-1-2=-3\end{align*}
• \begin{align*}f(1)=-3\end{align*}
• \begin{align*}\lim\limits_{x \to 1^+}f(x)=1^2-4=-3\end{align*}

Therefore, \begin{align*}\lim\limits_{x \to 1^-}f(x)=f(1)=\lim\limits_{x \to 1^+}f(x)\end{align*} and the function is continuous at \begin{align*}x=1\end{align*} and everywhere else.

#### Vocabulary

A one sided limit is a limit of a function when the evidence from only the positive or only the negative side is used to evaluate the limit.

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

#### Guided Practice

1. Megan argues that according to the definition of continuity, the following function is continuous. She says

• \begin{align*}\lim\limits_{x \to 2^-}f(x)=\infty\end{align*}
• \begin{align*}\lim\limits_{x \to 2^+}f(x)=\infty\end{align*}
• \begin{align*}f(2)=\infty\end{align*}

Thus since \begin{align*}\lim\limits_{x \to 2^-}f(x)=f(2)=\lim\limits_{x \to 2^+}f(x)\end{align*}, it meets the definition of continuous. How could you use the graph below to clarify Megan’s reasoning?

2. Evaluate the following limits.

1. \begin{align*}\lim\limits_{x \to 1^-}(2x-1)\end{align*}
2. \begin{align*}\lim\limits_{x \to -3^+}\left(\frac{2}{x+2}\right) \end{align*}
3. \begin{align*}\lim\limits_{x \to 2^+}\left(\frac{x^3-8}{x-2}\right) \end{align*}

3. Is the following function continuous?

\begin{align*}f(x)=\begin{cases} x^2 -1 \quad \ x < -1 \\ \quad 3 \quad \quad \ x=-1 \\ -x +3 \ -1< x \end{cases}\end{align*}

1. Megan is being extremely liberal with the idea of “\begin{align*}=\infty\end{align*}” because what she really means for the two limits is “DNE”. For the function evaluated at 2 the correct response is “undefined”. Two things that do not exist cannot be equal to one another.

2.

1.  \begin{align*}\lim\limits_{x \to 1^-}(2x-1)=2\cdot 1-1=2-1=1\end{align*}
2. \begin{align*}\lim\limits_{x \to -3^+}\left( \frac{2}{x+2}\right)=\frac{2}{-3+2}=\frac{2}{-1}=-2\end{align*}
3. \begin{align*}\lim\limits_{x \to 2^+}\left( \frac{x^3-8}{x-2}\right)=\lim\limits_{x \to 2^+}\left( \frac{(x-2)(x^2+2x+4)}{(x-2)}\right)=\lim\limits_{x \to 2^+}(x^2+2x+4)=2^2+2\cdot 2+4=12\end{align*}

3. Use the definition of continuity.

• \begin{align*}\lim\limits_{x \to 1^-}f(x)=(-1)^2-1=1-1=0\end{align*}
• \begin{align*}f(-1)=3\end{align*}
• \begin{align*}\lim\limits_{x \to -1^+}f(x)=-1+3=2\end{align*}

\begin{align*}\lim\limits_{x \to a^-}f(x)\ne f(a) \ne \lim\limits_{x \to a^+}f(x)\end{align*} so this function is discontinuous at \begin{align*}x=-1\end{align*}. It is continuous everywhere else.

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