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

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Although indeterminate forms usually show up when you are learning about L'Hôpital's Rule, you do not need to understand L'Hôpital's Rule to learn about indeterminate forms here. Just skip the references to L'Hôpital's Rule if you have not studied it yet.

The word 'indeterminate' means unable to determine. When you have an indeterminate form the value cannot be determinated. For example, \(0/0\) can mean \(1\) or \(-1227\) or \(-\infty\) or almost any number. There is no way to tell without changing the form (fascinating, isn't it?). That's where L'Hôpital's Rule comes in. L'Hôpital's Rule allows us to change the form into something else so that we CAN determine the value we are looking for when it comes to limits. Here is a table of most of the indeterminate forms you will run across.

Determinate-Indeterminate Forms Table

Indeterminate Forms

Determinate Forms

\( 0/0 \)

\( \infty + \infty = \infty \)

\( \pm \infty / \pm \infty \)

\( - \infty - \infty = - \infty\)

\( \infty - \infty \)

\( 0^{\infty} = 0 \)

\( 0 (\infty) \)

\( 0^{-\infty} = \infty \)

\( 0^0 \)

\( (\infty) \cdot (\infty) = \infty \)

\( 1^{\infty} \)

\( \infty ^ 0 \)

Use L'Hôpital's Rule

Do Not Use L'Hôpital's Rule

Types of Indeterminate Forms

From the discussion so far, it looks like we have a lot going on and there are a lot of cases in the indeterminate forms table. However, there are really only four cases and only two techniques you need to cover all the cases. Let's look at the cases.

Type 1 - - Indeterminate Quotients
In a perfect world, all your problems would be in this form because you can just apply L'Hôpital's Rule directly. These are problems where you already have a limit in the form
\(\displaystyle{ \lim_{x \to a}{\frac{n(x)}{d(x)}} }\)
And, when you plug in \(x=a\), you get either \(0/0\) or \(\pm \infty / \pm \infty\). In the last indeterminate form with \(\infty\), the signs can be anything, so you have 4 possible cases. In any case, your problem is all set up to use L'Hôpital's Rule and you just directly apply the rule.

Type 2 - - Indeterminate Products
In this case, you have an determinate form that looks like \(0(\infty)\) or \(0(-\infty)\), in which case you just need to do a little bit of algebra to convert one of the pieces of the product into a denominator and you end up with case 1. This sometimes takes some thinking. For example, if you have
\(\displaystyle{ \lim_{x \to 0^+}{x\ln(x)} }\), you can rewrite \(x\) as \(1/x^{-1}\) and the limit now is
\(\displaystyle{ \lim_{x \to 0^+}{\ln(x)/x^{-1}} }\) where \(n(x) = \ln(x)\) and \(d(x) = x^{-1}\) and you have the indeterminate form \(\infty / \infty\) and you can use L'Hôpital's Rule.

Type 3 - - Indeterminate Differences
In this case, you have the indeterminate form that looks like \(\infty - \infty\) and we need a fraction to be able to apply L'Hôpital's Rule. To get the right form, just find a common a denominator and combine the terms to get an indeterminate form that looks like type 1.

Type 4 - - Indeterminate Powers
In the previous three types, you either started with a fraction in the right form (type 1) or did some algebra to get your function into the right form (type 2 and 3). In this case, you need another technique. The idea is that you have an indeterminate form that looks like \( 0^0 \), \( 1^{\infty} \) or \( \infty ^ 0 \) and you need to get some kind of fraction in order to apply L'Hôpital's Rule. To do this we use the natural logarithm. Let's go through a general example. We have the limit \(\displaystyle{ \lim_{x \to a}{[f(x)]^{g(x)}} }\) and \([f(a)]^{g(a)}\) is an indeterminate power.
Set \(y = [f(x)]^{g(x)}\) and take the natural log of both sides.
So that \(\ln(y) = \ln( [f(x)]^{g(x)} ) \) and using the power rule for logarithms, we have \(\ln(y) = g(x) \ln[ f(x) ] \).
We take the limit of \(\ln(y)\) using one types 1-3 above to get a value \(L\), which may be finite or infinite.
To get the final answer, we 'undo' the natural log by putting \(L\) in the exponent of \(e\) giving \(e^L\).

This discussion in the cases above may be a bit confusing, especially case 4, but, if you look at the second example coming up, we will use this idea to show that \(1^{\infty}\) is indeterminate.

The indeterminate forms table above shows most of the indeterminate forms that you will run across. However, you do not need to memorize these forms to determine if you need to use L'Hôpital's Rule. There is an easier way. All you need to remember is that indeterminate form is \(0/0\) and determinate form is \(c/0 = \pm \infty\) where \( c \neq 0 \). Then use the ideas in one of the above cases to get the fraction \(0/0\). Let's do some examples.

Example 1

Determine if \(\displaystyle{ \infty / \infty }\) is determinate or indeterminate.

Solution

Remember that \( 1/0 = \infty \). So we will replace \(\infty\) in the equation. So now we have \(\displaystyle{ \frac{\infty}{\infty} = \frac{1/0}{1/0} = \frac{1}{0}\frac{0}{1} = \frac{1}{1}\frac{0}{0} = \frac{0}{0} }\).
The last fraction is indeterminate, so \(\displaystyle{ \frac{\infty}{\infty} }\) is also indeterminate.

Comments:
1. In a fraction, you can cancel non-zero numbers (\(1\) in this example) but you can never cancel zeros since a zero in the denominator is a problem.
2. \(1/0\) is just one possible form of \(\infty\). You could have used \(3/0\) or \(-17/0\) or just about any non-zero value in the numerator. They will cancel in the end.

Example 2

Determine if \( 1^{\infty}\) is determinate or indeterminate.

Solution

In this case, we need to get this in fraction form to compare it to \(0/0\). Remember your logarithms? We will use the natural logarithm since it is the most common logarithm in calculus.
\(\begin{array}{rcl} y & = & 1^{\infty} \\ \ln(y) & = & \ln \left[ 1^{\infty} \right] \\ & = & \infty (\ln 1) \\ & = & (1/0) (0) = 0/0 \end{array}\)
Since the last fraction is indeterminate, \( 1^{\infty} \) is also indeterminate.

Comments:
1. We are being a bit 'flexible' with logarithms here but, for the purposes of this example, the work is accurate enough to support the answer.
2. You may have noticed that we ended up with \( \ln(y) = 0/0 \) rather than \( y = 0/0 \). The conclusion still holds.

Now that you have some guidelines, try showing these on the other items in the indeterminate forms table. This will not only give you practice, but it will help you remember the techniques quickly at exam time.

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Topics You Need To Understand For This Page

Related Topics and Links

external links you may find helpful

Wikipedia - Indeterminate Form

Trig Formulas

The Unit Circle

The Unit Circle [wikipedia]

Basic Trig Identities

Set 1 - basic identities

\(\displaystyle{ \tan(t) = \frac{\sin(t)}{\cos(t)} }\)

\(\displaystyle{ \cot(t) = \frac{\cos(t)}{\sin(t)} }\)

\(\displaystyle{ \sec(t) = \frac{1}{\cos(t)} }\)

\(\displaystyle{ \csc(t) = \frac{1}{\sin(t)} }\)

Set 2 - squared identities

\( \sin^2t + \cos^2t = 1\)

\( 1 + \tan^2t = \sec^2t\)

\( 1 + \cot^2t = \csc^2t\)

Set 3 - double-angle formulas

\( \sin(2t) = 2\sin(t)\cos(t)\)

\(\displaystyle{ \cos(2t) = \cos^2(t) - \sin^2(t) }\)

Set 4 - half-angle formulas

\(\displaystyle{ \sin^2(t) = \frac{1-\cos(2t)}{2} }\)

\(\displaystyle{ \cos^2(t) = \frac{1+\cos(2t)}{2} }\)

Trig Derivatives

\(\displaystyle{ \frac{d[\sin(t)]}{dt} = \cos(t) }\)

 

\(\displaystyle{ \frac{d[\cos(t)]}{dt} = -\sin(t) }\)

\(\displaystyle{ \frac{d[\tan(t)]}{dt} = \sec^2(t) }\)

 

\(\displaystyle{ \frac{d[\cot(t)]}{dt} = -\csc^2(t) }\)

\(\displaystyle{ \frac{d[\sec(t)]}{dt} = \sec(t)\tan(t) }\)

 

\(\displaystyle{ \frac{d[\csc(t)]}{dt} = -\csc(t)\cot(t) }\)

Inverse Trig Derivatives

\(\displaystyle{ \frac{d[\arcsin(t)]}{dt} = \frac{1}{\sqrt{1-t^2}} }\)

 

\(\displaystyle{ \frac{d[\arccos(t)]}{dt} = -\frac{1}{\sqrt{1-t^2}} }\)

\(\displaystyle{ \frac{d[\arctan(t)]}{dt} = \frac{1}{1+t^2} }\)

 

\(\displaystyle{ \frac{d[\arccot(t)]}{dt} = -\frac{1}{1+t^2} }\)

\(\displaystyle{ \frac{d[\arcsec(t)]}{dt} = \frac{1}{\abs{t}\sqrt{t^2 -1}} }\)

 

\(\displaystyle{ \frac{d[\arccsc(t)]}{dt} = -\frac{1}{\abs{t}\sqrt{t^2 -1}} }\)

Trig Integrals

\(\int{\sin(x)~dx} = -\cos(x)+C\)

 

\(\int{\cos(x)~dx} = \sin(x)+C\)

\(\int{\tan(x)~dx} = -\ln\abs{\cos(x)}+C\)

 

\(\int{\cot(x)~dx} = \ln\abs{\sin(x)}+C\)

\(\int{\sec(x)~dx} = \) \( \ln\abs{\sec(x)+\tan(x)}+C\)

 

\(\int{\csc(x)~dx} = \) \( -\ln\abs{\csc(x)+\cot(x)}+C\)

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Meaning of Indeterminate

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