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 | ||
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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 |
Type 2 - - Indeterminate Products |
Type 3 - - Indeterminate Differences |
Type 4 - - Indeterminate Powers |
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 |
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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 \(\displaystyle{ \frac{\infty}{\infty} = \frac{1/0}{1/0} = \frac{1}{0}\frac{0}{1} = \frac{1}{1}\frac{0}{0} = \frac{0}{0} }\). |
Example 2 |
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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. |
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.
You CAN Ace Calculus
external links you may find helpful |
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The Unit Circle
The Unit Circle [wikipedia]
Basic Trig Identities
Set 1 - basic identities | |||
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\(\displaystyle{ \tan(t) = \frac{\sin(t)}{\cos(t)} }\) |
\(\displaystyle{ \cot(t) = \frac{\cos(t)}{\sin(t)} }\) |
\(\displaystyle{ \sec(t) = \frac{1}{\cos(t)} }\) |
\(\displaystyle{ \csc(t) = \frac{1}{\sin(t)} }\) |
Set 2 - squared identities | ||
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\( \sin^2t + \cos^2t = 1\) |
\( 1 + \tan^2t = \sec^2t\) |
\( 1 + \cot^2t = \csc^2t\) |
Set 3 - double-angle formulas | |
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\( \sin(2t) = 2\sin(t)\cos(t)\) |
\(\displaystyle{ \cos(2t) = \cos^2(t) - \sin^2(t) }\) |
Set 4 - half-angle formulas | |
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\(\displaystyle{ \sin^2(t) = \frac{1-\cos(2t)}{2} }\) |
\(\displaystyle{ \cos^2(t) = \frac{1+\cos(2t)}{2} }\) |
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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