\( \newcommand{\abs}[1]{\left| \, {#1} \, \right| } \) \( \newcommand{\cm}{\mathrm{cm} } \) \( \newcommand{\sec}{ \, \mathrm{sec} \, } \) \( \newcommand{\units}[1]{\,\text{#1}} \) \( \newcommand{\vhat}[1]{\,\hat{#1}} \) \( \newcommand{\vhati}{\,\hat{i}} \) \( \newcommand{\vhatj}{\,\hat{j}} \) \( \newcommand{\vhatk}{\,\hat{k}} \) \( \newcommand{\vect}[1]{\boldsymbol{\vec{#1}}} \) \( \newcommand{\norm}[1]{\|{#1}\|} \) \( \newcommand{\arccot}{ \, \mathrm{arccot} \, } \) \( \newcommand{\arcsec}{ \, \mathrm{arcsec} \, } \) \( \newcommand{\arccsc}{ \, \mathrm{arccsc} \, } \) \( \newcommand{\sech}{ \, \mathrm{sech} \, } \) \( \newcommand{\csch}{ \, \mathrm{csch} \, } \) \( \newcommand{\arcsinh}{ \, \mathrm{arcsinh} \, } \) \( \newcommand{\arccosh}{ \, \mathrm{arccosh} \, } \) \( \newcommand{\arctanh}{ \, \mathrm{arctanh} \, } \) \( \newcommand{\arccoth}{ \, \mathrm{arccoth} \, } \) \( \newcommand{\arcsech}{ \, \mathrm{arcsech} \, } \) \( \newcommand{\arccsch}{ \, \mathrm{arccsch} \, } \)

17Calculus - One-Sided Limits

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One-sided limits require a good understanding of piecewise functions. If you are a little rusty or just need a quick reminder, you can find a complete discussion of piecewise functions on the piecewise functions precalculus page.

One-sided limits are finite limits where we evaluate the limit from each side of a point individually. The notation we use is

\(\displaystyle{ \lim_{x \to a^-}{f(x)} }\)

evaluate the limit on the left side of \(a\), i.e. values of \(x < a\)

\(\displaystyle{ \lim_{x \to a^+}{f(x)} }\)

evaluate the limit on the right side of \(a\), i.e. values of \(x > a\)

The negative and positive sign that look like exponents on the finite value \(a\) indicate the side that we are looking at.

One of the reasons we need to look at limits on both sides of some number is when we are determining continuity. As you know from the continuity page, one of the requirements for continuity is that the limit at a point must exist. In order for a limit to exist, the limit from the left must be equal to the limit from the right, i.e. \(\displaystyle{\lim_{x \to a^-}{f(x)} = \lim_{x \to a^+}{f(x)}}\).

Notice that we are NOT saying that the function value must be equal to the limit or even that the function need be defined at \(x=a\), only that the limit be equal on both sides of a.

Here is a great video to build your intuition of one-sided limits. He uses an absolute value function to discuss the idea of one-sided limits and limits that do not exist.

Dr Chris Tisdell - Limit of a function [9min-11secs]

video by Dr Chris Tisdell

Here is a good video showing a graph with several one-sided limits.

Krista King Math - How to find limits on CRAZY GRAPHS [7min-47secs]

video by Krista King Math

You will see more one-sided limits when you learn about continuity. For now, let's work these practice problems.

Calculus, Better Explained: A Guide To Developing Lasting Intuition

Practice

Unless otherwise instructed, evaluate these limits. Give your answers in exact form.

Basic

\(\displaystyle{\lim_{x\to1}{\frac{|x-1|}{x-1}}}\)

Problem Statement

Evaluate the limit, giving your answer in exact form. \(\displaystyle{\lim_{x\to1}{\frac{|x-1|}{x-1}}}\)

Solution

PatrickJMT - 484 video solution

video by PatrickJMT

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\(\displaystyle{\lim_{x\to5^+}{\frac{6}{x-5}}}\)

Problem Statement

\(\displaystyle{\lim_{x\to5^+}{\frac{6}{x-5}}}\)

Final Answer

\(\displaystyle{\lim_{x\to5^+}{\frac{6}{x-5}}}\) \( = + \infty \)

Problem Statement

\(\displaystyle{\lim_{x\to5^+}{\frac{6}{x-5}}}\)

Solution

Okay, so first if we try to just evaluate the limit at \(x=5\), we get \(6/0\). Our first inclination might be to just say the limit is positive infinity. However \(6/0\) doesn't give us enough information to say whether the limit goes to positive infinity or negative infinity. We need to know the sign of the denominator.
As the denominator gets smaller and smaller, the magnitude of the fraction gets larger and larger but, since the numerator is positive, the sign of denominator will determine whether the limit goes to positive or negative infinity.

Here is the logic we use to determine the limit. Since \(x\) is approaching \(5\) from the right, \(x\) will always be larger than \(5\), so \(x-5\) is always positive. To check this, let's test a few numbers.
\(x=5.1 \to 6/(5.1-5) = 6/0.1 = 60 \gt 0 \)
\(x = 5.01 \to 6/(5.01-5) = 6/0.01 = 600 \gt 0\)

So the test values go along with our argument about the denominator always being positive. And since the numerator is also always positive, then the resulting fraction will always be positive. So now we can confidently say that the limit goes to positive infinity.

Although we were not asked to in the problem, we decided to plot the function to check our answer. Here is the plot.

Final Answer

\(\displaystyle{\lim_{x\to5^+}{\frac{6}{x-5}}}\) \( = + \infty \)

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\(\displaystyle{\lim_{x\to0}{\frac{x-1}{x^2(x+2)}}}\)

Problem Statement

\(\displaystyle{\lim_{x\to0}{\frac{x-1}{x^2(x+2)}}}\)

Final Answer

\(\displaystyle{\lim_{x\to0}{\frac{x-1}{x^2(x+2)}}}\) \( = - \infty \)

Problem Statement

\(\displaystyle{\lim_{x\to0}{\frac{x-1}{x^2(x+2)}}}\)

Solution

Okay, so first if we try to just evaluate the limit at \(x=0\), we get \(-1/0\). Our first inclination might be to just say the limit is negative infinity. However \(-1/0\) doesn't give us enough information to say whether the limit goes to positive infinity or negative infinity. We need to know the signs of both the numerator and denominator on both sides of zero.
First, let's look at the denominator. Of course, the \(x^2\) term is always positive. Since \(x\) is very close to zero, \(x+2\) also always positive. This is the case whether \(x\) is positive or negative. Consequently, the sign of the numerator will determine the limit.

Here is the logic we use to determine the limit. We need to look at each side of \(x=0\) separately.

For \(x \lt 0 \), \(x-1\) is negative. Similarly for \(x \gt 0\), for \(x\) very close to zero, \(x-1\) is also negative. In equation form, \(\displaystyle{ \lim_{x\to0^-}{ \frac{x-1}{x^2(x+2)} } = - \infty }\) and \(\displaystyle{ \lim_{x\to0^+}{ \frac{x-1}{x^2(x+2)} } = - \infty }\). Since the limit from the left is equal to the limit from the right, we know that \(\displaystyle{ \lim_{x\to0}{ \frac{x-1}{x^2(x+2)} } = - \infty }\).

Although we were not asked to in the problem, we decided to plot the function to check our answer. Here is the plot.

Final Answer

\(\displaystyle{\lim_{x\to0}{\frac{x-1}{x^2(x+2)}}}\) \( = - \infty \)

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\(\displaystyle{\lim_{x\to-4^-}{|x+4|}}\)

Problem Statement

Evaluate the following limit, giving your answer in exact form. \(\displaystyle{\lim_{x\to-4^-}{|x+4|}}\)

Solution

PatrickJMT - 489 video solution

video by PatrickJMT

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\(\displaystyle{\lim_{x\to5^+}{\sqrt{x^2-25}}}\)

Problem Statement

Evaluate the limit, giving your answer in exact form. \(\displaystyle{\lim_{x\to5^+}{\sqrt{x^2-25}}}\)

Solution

PatrickJMT - 490 video solution

video by PatrickJMT

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\(\displaystyle{\lim_{x\to5^-}{\sqrt{x(5-x)}}}\)

Problem Statement

Evaluate the limit, giving your answer in exact form. \(\displaystyle{\lim_{x\to5^-}{\sqrt{x(5-x)}}}\)

Solution

Krista King Math - 491 video solution

video by Krista King Math

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\(\displaystyle{ \lim_{x\to0}{1/x} }\)

Problem Statement

Evaluate the limit \(\displaystyle{ \lim_{x\to0}{1/x} }\) giving your answer in exact form.

Final Answer

The limit does not exist.

Problem Statement

Evaluate the limit \(\displaystyle{ \lim_{x\to0}{1/x} }\) giving your answer in exact form.

Solution

For this solution, we will look at specific values of \(x\) to give us a feel for what is happening. Although not recommended for most problems, this technique, along with the graph, can give a good understanding of this function around \(x=0\). Of course, we can't just plug in zero for \(x\) since we end up with \(1/0\) which is not valid.

Here is the plot of \(1/x\).

First, let's look at values of \(f(x)=1/x\) for \(x \gt 0\).

\( f(2) = 1/2 \)

\( f(1) = 1 \)

\( f(0.1) = 10 \)

\( f(0.01) = 100 \)

\( f(0.001) = 1000 \)

So it looks like as \(x\) gets smaller, \(f(x)\) gets larger. This can be seen in the graph. So on the right side of the graph, \(\displaystyle{ \lim_{x\to0^+}{1/x} = \infty }\).

Now let's look at values of \(f(x)=1/x\) for \(x \lt 0\).

\( f(-2) = -1/2 \)

\( f(-1) = -1 \)

\( f(-0.1) = -10 \)

\( f(-0.01) = -100 \)

\( f(-0.001) = -1000 \)

These values are getting smaller, which you can also see in the graph. So on the left side of the graph \(\displaystyle{ \lim_{x\to0^-}{1/x} = -\infty }\).

Since the limits are different on each side of the graph, i.e. \(\displaystyle{ \lim_{x\to0^+}{1/x} \neq \lim_{x\to0^-}{1/x} }\), the limit \(\displaystyle{\lim_{x\to0}{1/x}}\) does not exist.

Final Answer

The limit does not exist.

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\(\displaystyle{ \lim_{x\to0}{1/x^2} }\)

Problem Statement

Evaluate the limit \(\displaystyle{ \lim_{x\to0}{1/x^2} }\) giving your answer in exact form.

Final Answer

\(\displaystyle{ \lim_{x\to0}{1/x^2} }\) \( = \infty \)

Problem Statement

Evaluate the limit \(\displaystyle{ \lim_{x\to0}{1/x^2} }\) giving your answer in exact form.

Solution

For this solution, we will look at specific values of \(x\) to give us a feel for what is happening. Although not recommended for most problems, this technique, along with the graph, can give a good understanding of this function around \(x=0\). Of course, we can't just plug in zero for \(x\) since we end up with \(1/0\) which is not valid.

Here is the plot of \(1/x^2\).

First, let's look at values of \(f(x)=1/x^2\) for \(x \gt 0\).

\( f(2) = 1/4 \)

\( f(1) = 1 \)

\( f(0.1) = 100 \)

\( f(0.01) = 10000 \)

\( f(0.001) = 1000000 \)

So it looks like as \(x\) gets smaller, \(f(x)\) gets larger. This can be seen in the graph. So on the right side of the graph, \(\displaystyle{ \lim_{x\to0^+}{1/x^2} = \infty }\).

Now let's look at values of \(f(x)=1/x\) for \(x \lt 0\).

\( f(-2) = 1/4 \)

\( f(-1) = 1 \)

\( f(-0.1) = 100 \)

\( f(-0.01) = 10000 \)

\( f(-0.001) = 1000000 \)

These values are also getting larger, which you can also see in the graph. So on the left side of the graph \(\displaystyle{ \lim_{x\to0^-}{1/x} = \infty }\).

Since \(\displaystyle{ \lim_{x\to0^+}{1/x} = \lim_{x\to0^-}{1/x} }\), the limit \(\displaystyle{\lim_{x\to0}{1/x} = \infty}\).

Final Answer

\(\displaystyle{ \lim_{x\to0}{1/x^2} }\) \( = \infty \)

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Intermediate

Evaluate \(\displaystyle{\lim_{x\to1}{f(x)}}\) for \(\displaystyle{f(x)=\left\{\begin{array}{lr} x+3 & x \leq 1 \\ x^2-2x & x >1 \end{array}\right.}\)

Problem Statement

Evaluate \(\displaystyle{\lim_{x\to1}{f(x)}}\) for \(\displaystyle{f(x)=\left\{\begin{array}{lr} x+3 & x \leq 1 \\ x^2-2x & x >1 \end{array}\right.}\)

Solution

PatrickJMT - 488 video solution

video by PatrickJMT

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Prove that the limit \(\displaystyle{\lim_{x\to0}{\frac{|x|}{x}}}\) does not exist.

Problem Statement

Prove that the limit \(\displaystyle{\lim_{x\to0}{\frac{|x|}{x}}}\) does not exist.

Solution

Krista King Math - 492 video solution

video by Krista King Math

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\(\displaystyle{\lim_{x\to0}{\frac{x-2\abs{x}}{\abs{x}}}}\)

Problem Statement

Evaluate the limit \(\displaystyle{\lim_{x\to0}{\frac{x-2\abs{x}}{\abs{x}}}}\) giving your answer in exact form.

Final Answer

The limit does not exist.

Problem Statement

Evaluate the limit \(\displaystyle{\lim_{x\to0}{\frac{x-2\abs{x}}{\abs{x}}}}\) giving your answer in exact form.

Solution

This is a quite interesting function. First, let's look at the equation. Since the function \(\displaystyle{ f(x) = \frac{x-2\abs{x}}{\abs{x}} }\) contains absolute values, we need to set up a piecewise function.

When \(x \lt 0\), \(\displaystyle{ f(x) = \frac{x+2x}{-x} = -3 }\)

When \(x \gt 0\), \(\displaystyle{ f(x) = \frac{x-2x}{x} = -1 }\)

So our piecewise function is
\( f(x) = \left\{ \begin{array}{rrl} -3 & & x \lt 0 \\ -1 & & x \gt 0 \end{array} \right. \)

Here is the plot of \(f(x)\).

So we can see from the piecewise function and the plot that \(\displaystyle{\lim_{x\to0^+}{f(x)} = -1 }\) and \(\displaystyle{\lim_{x\to0^-}{f(x)} = -3 }\). Since the limits are different on each side of \(x=0\), the limit does not exist.

Final Answer

The limit does not exist.

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Unless otherwise instructed, evaluate these limits. Give your answers in exact form.

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