Dividing by zero is considered a mathematical taboo because the operation itself does not make sense. In this case, if the numerator is other than zero, then we say that the operation is undefined.
There are other types of operations that you might find that is also problematic. Can you divide \(0\) by \(0\)? The answer is that…not quite. However, you can find the limit of the quotient of two numbers as both approach zero. Surprisingly enough, you can have different answers depending on how this division is approached. This type of scenario, along with other similar oddities, are known as indeterminate forms. Here, you will learn how to deal with them.
Definition of Indeterminate Forms
Begin by recalling what an indeterminate form is.
An indeterminate form is an expression of two functions whose limit cannot be evaluated by direct substitution.
The most common indeterminate forms are:
\[ \frac{0}{0}\]
and
\[ \frac{\pm \infty}{ \pm \infty}\]
The above indeterminate forms are typically solved using L'Hôpital's rule, as they are already written in the way you require for the rule to work.
It is crucial that you have a grasp on what L'Hôpital's rule is and how to use it to evaluate limits. If you need a refresher, please reach out to our related articles.
However, these are not the only indeterminate forms. There are more indeterminate forms, which are usually addressed as the other indeterminate forms.
The other indeterminate forms are the following:
- \( \infty \cdot 0\)
- \(0^0\)
- \(1^\infty\)
- \(\infty-\infty\)
- \(\infty^0\)
These indeterminate forms can also be solved using L'Hôpital's rule, but as the rule requires rational expressions, you will need to do a bit of algebra before applying the rule.
Indeterminate Forms of Limits
As you just found previously, you will find indeterminate forms whenever you are trying to evaluate limits by direct substitution.
Consider the following limit:
\[ \lim_{x \to 4} \frac{x^2-16}{x-4}\]
If you try to substitute \(x\) for \(4\) in the above limit, you will find that:
\[ \begin{align} \lim_{x \to 4} \frac{x^2-16}{x-4} &= \frac{4^2-16}{4-4} \\ &= \frac{16-16}{4-4} \\ &= \frac{0}{0} \end{align}\]
which is an indeterminate form of:
\[ \frac{0}{0}\]
To properly evaluate this limit, you can factor the difference of squares, so you can cancel the like terms, that is:
\[ \begin{align} \lim_{x \to 4} \frac{x^2-16}{x-4} &= \lim_{x \to 4} \frac{(x+4)\cancel{(x-4)}}{\cancel{(x-4)}} \\ &= \lim_{x \to 4} (x+4) \\ &= 4+4 \\&= 8\end{align}\]
Whenever you try to evaluate a limit by direct substitution just to find out any of the above operations involving \(0\) or infinity, then you are dealing with an indeterminate form.
Indeterminate Forms and L'Hôpital's Rule
Sometimes, you will find that the involved limit cannot be simplified in any way, or maybe the simplification just does not come to your mind. In this case, you can use L'Hôpital's rule.
L'Hôpital's rule is a method for evaluating limits that result in indeterminate forms.
L'Hôpital's rule tells you that, if a limit of the quotient of two functions evaluates to an indeterminate form, then:\[ \lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a}\frac{f'(x)}{g'(x)}\]
Provided that the limits exist.
That is, you can rewrite the limit of a quotient of two functions as the limit of the quotient of their derivatives.
If the limit does not result in an indeterminate form, you cannot use L'Hôpital's rule!
This becomes particularly useful because functions like power functions tend to become simpler as you differentiate them.
In the previous example, you evaluated the limit:
\[ \lim_{x \to 4} \frac{x^2-16}{x-4}\]
By factorizing the numerator. If this particular factorization does not come to your mind, you can also use L'Hôpital's rule, obtaining:
\[ \begin{align} \lim_{x \to 4} \frac{x^2-16}{x-4} &= \lim_{x \to 4} \frac{2x}{1} \\ &= \frac{2(4)}{1} \\ &= 8\end{align} \]
Indeterminate Forms That Involve Infinity
Most (but not all) indeterminate forms involve infinity in some way. You can categorize indeterminate forms based on which operation is being indeterminate.
Subtraction
One of the other indeterminate forms you will find is
\[ \infty - \infty\]
These expressions typically appear when adding or subtracting rational expressions, so it is advised that you work out the fractions and simplify them as much as possible.
Evaluate the limit:
\[ \lim_{x \to 0^+} \left( \frac{1}{x}-\frac{1}{x^2} \right)\]
Solution:
If you were to evaluate the limit by direct substitution, you would find that:
\[ \lim_{ x \to 0^+} \left( \frac{1}{x} - \frac{1}{x^2}\right)= \infty - \infty\]
So this is an indeterminate form.
Instead of evaluating directly, try subtracting both fractions, that is:
\[ \lim_{ x \to 0^+} \left( \frac{1}{x}-\frac{1}{x^2} \right)= \lim_{x \to 0^+} \left( \frac{x-1}{x^2}\right)\]
So you can inspect the limit by direct substitution.
Remember that, in oder to use L'Hôpital's rule, you need to have an indeterminate form of \( 0/0\) or \(\infty/\infty\). Always inspect the limit first by direct substitution.
The numerator is negative, and the denominator is positive as you approach \(0\) from the right, so the result will be negative infinity, that is:
- An indeterminate form is an expression of two functions whose limit cannot be evaluated by direct substitution.
- The most common indeterminate forms are \(0/0\) and \( \pm\infty/\pm\infty\).
- The other indeterminate forms refer to the expressions \(0 \cdot \infty\), \(0^0\), \( \infty^0\), \(1^\infty\), and \(\infty-\infty\).
- A limit resulting in an indeterminate form that involves quotients can be evaluated using L'Hôpital's Rule.
- By algebraic means, it is possible to transform limits that involve the other indeterminate forms into expressions that involve quotients, so you can use L'Hôpital's rule to evaluate them.
\[ \lim_{x \to 0^+} \left( \frac{1}{x}-\frac{1}{x^2} \right) = -\infty\]
After subtracting (or, in some scenarios, adding) the fractions, you will be left with a rational expression, so you can use L'Hôpital's rule if the limit does not evaluate directly.
Evaluate the limit
\[ \lim_{x \to 0^+} \left( \frac{\cos{x}}{x}-\frac{1}{x}\right)\]
Solution:
Once again, if you were to evaluate the limit directly, you would find that:
\[ \lim_{x \to 0^+} \left( \frac{\cos{x}}{x}-\frac{1}{x}\right) = \infty-\infty\]
By subtracting the fractions, you get:
\[ \lim_{x \to 0^+} \left( \frac{\cos{x}}{x}-\frac{1}{x}\right) = \lim_{x \to 0^+} \frac{\cos{x}-1}{x}\]
The cosine of \(0\) is \(1,\) so both the numerator and the denominator approach \(0\) as \(x \to 0.\) This suggests the use of L'Hôpital's rule, that is:
\[ \lim_{x \to 0^+} \left( \frac{\cos{x}}{x}-\frac{1}{x}\right) = \lim_{x \to 0^+}\frac{\sin{x}}{1}\]
Since the sine of \(0\) is \(0\), you can now evaluate the limit, obtaining:
\[ \lim_{x \to 0^+} \left( \frac{\cos{x}}{x}-\frac{1}{x}\right) =0\]
Multiplication
This indeterminate form comes as the expression
\[ 0 \cdot \infty.\]
You cannot use L'Hôpital's rule because of the product of two functions, so all you need to do is to rewrite the product as a fraction by recalling that
\[ f(x) \cdot g(x) = f(x) \cdot \frac{1}{\frac{1}{g(x)}}.\]
Let
\[ h(x) = \frac{1}{g(x)},\]
so
\[ \begin{align} f(x) \cdot g(x) &= f(x) \cdot \frac{1}{h(x)} \\ &= \frac{f(x)}{h(x)}. \end{align}\]
This means that you can now use L'Hôpital's rule!
Evaluate the limit
\[ \lim_{x \to \infty} x\, e^{-x}.\]
Solution:
As usual, begin by trying to evaluate the limit directly. The limit as \(x \to \infty\) of \(e^{-x}\) is \(0\), so you are dealing with an indeterminate form of \( \infty \cdot 0\). To use L'Hôpital, note that you can write \(e^{-x}\) as \(e^x\) in the denominator, that is
\[ \lim_{x \to \infty} x\,e^{-x} = \lim_{x \to \infty}\frac{x}{e^x}.\]
Now you have an indeterminate form of \( \infty/\infty\), so use L'Hôpital's rule,
\[ \begin{align} \lim_{x \to \infty} x\,e^{-x} &= \lim_{x \to \infty} \frac{1}{e^x}\ \\ &= 0. \end{align}\]
Exponentiation
These indeterminate forms come as
\[ 0^\infty,\]
\[ 0^0,\]
and
\[1^\infty.\]
You can use the properties of logarithms to address any of the above indeterminate forms. Consider the case
\[ f(x)^{g(x)}.\]
By using the natural logarithm, you can find that
\[ \ln{ \left( f(x)^{g(x)}\right)} = g(x) \ln{\left( f(x) \right)},\]
which means that you can transform exponentiation into a product by using the natural logarithm.
The limits that result in any of the above indeterminate forms typically come as
\[ \lim_{x \to a} f(x)^{g(x)}.\]
Because the natural logarithmic function is a continuous function, you can evaluate the natural logarithm of the limit, and then undo the natural logarithm by using the exponential function.
Evaluate the limit
\[\lim_{x \to 0^+} x^x.\]
Solution:
Begin by labeling the limit, namely
\[ L = \lim_{x \to 0^+}x^x.\]
From here, you can take the natural logarithm of both sides, that is
\[ \ln{L}=\ln{\left( \lim_{x \to 0^+} x^x \right)}.\]
Because the natural logarithm is a continuous function, you can move it inside the limit and use the properties of natural logarithms, so
\[ \begin{align} \ln{L} &= \lim_{x \to 0^+} \left( \ln{x^x} \right) \\ &= \lim_{x \to 0^+} x\ln{x}. \end{align}\]
As \(x \to 0^+\), the natural logarithm goes to negative infinity, so the above expression is an indeterminate form of \(0 \cdot \infty\), which you can work using some algebra
\[ \begin{align} \ln{L} &= \lim_{x \to 0^+} x\ln{x} \\ &= \lim_{x \to 0^+} \frac{\ln{x}}{\frac{1}{x}}, \end{align}\]
and then use L'Hôpital's rule, so
\[ \begin{align} \ln{L} &= \lim_{x \to 0^+} \frac{\frac{1}{x}}{-\frac{1}{x^2}} \\ &= \lim_{x \to 0^+} (-x) \\ &= 0. \end{align} \]
Finally, undo the natural logarithm by using the exponential function, so
\[ \begin{align} L &= e^0 \\ &= 1. \end{align}\]
Examples of Indeterminate Forms
Try working on more examples to be proficient in evaluating the limits of indeterminate forms!
Evaluate the limit
\[\lim_{x \to 0^+} \left(\frac{1}{x}-\csc{x} \right).\]
Solution:
Begin by recalling that the cosecant function is the reciprocal of the sine function, so
\[ \lim_{ x \to 0^+} \left( \frac{1}{x}-\csc{x} \right) = \lim_{x \to 0^+} \left( \frac{1}{x}-\frac{1}{\sin{x}}\right).\]
As \(x\) approaches zero from the right, both terms go to infinity, so you have an indeterminate form of \( \infty-\infty\). Work this around by subtracting the fractions
\[ \lim_{ x \to 0^+} \left( \frac{1}{x}-\frac{1}{\sin{x}}\right) = \lim_{x \to 0^+} \left( \frac{\sin{x}-x}{x\sin{x}}\right),\]
which is now an indeterminate form of \(0/0\). Use L'Hôpital's rule, that is
\[ \lim_{ x \to 0^+} \left( \frac{1}{x}-\frac{1}{\sin{x}}\right) = \lim_{x \to 0^+} \frac{\cos{x}-1}{\sin{x}+x\cos{x}},\]
The derivative of \(x\sin{x}\) is \(\sin{x}+x\cos{x}\).
but you will find that this is another indeterminate form of \(0/0\). Use L'Hôpital's rule once more, so
\[ \lim_{ x \to 0^+} \left( \frac{1}{x}-\frac{1}{\sin{x}}\right) = \lim_{x \to 0^+} \frac{\sin{x}}{\cos{x}+\cos{x}-x\sin{x}},\]
The derivative of \(x\cos{x}\) is \(\cos{x}-x\sin{x}\).
which can now be evaluated, giving you
\[ \lim_{ x \to 0^+} \left( \frac{1}{x}-\frac{1}{\sin{x}}\right) =0.\]
Here is an example involving the product of zero and infinity.
Evaluate the limit
\[ \lim_{x \to 0^+} x\cot{x}.\]
Solution:
Remember that the cotangent function is the reciprocal of the tangent function. Since \(\tan{0}=0\), the cotangent goes to infinity when approached from the right, so this is an indeterminate form of \(0 \cdot \infty.\) To solve this, rewrite the cotangent as the reciprocal of the tangent, that is
\[ \lim_{x \to 0^+} x\cot{x} = \lim_{x \to 0^+} \frac{x}{\tan{x}},\]
which is now an indeterminate form of \(0/0\), so use L'Hôpitals rule
\[ \lim_{x \to 0^+} x\cot{x} = \lim_{x \to 0^+} \frac{1}{\sec^2{x}}.\]
The secant of \(0\) is equal to \(1\), so
\[ \lim_{x \to 0^+} x\cot{x}=1.\]
It is time for an exponentiation one.
Evaluate the limit
\[ \lim_{x \to \infty}x^{^1/_x}. \]
Solution:
As \(x\) goes to infinity, \(1/x\) goes to zero, so this is an indeterminate form of \(\infty^0\). Label the limit as \(L\) and find its natural logarithm, that is
\[ \ln{L} = \ln{\left( \lim_{x \to \infty} x^{^1/_x} \right)}, \]
and use the fact that the natural logarithm is a continuous function to introduce it inside the limit, so
\[ \ln{L} = \lim_{ x\to \infty} \ln{\left( x^{^1/_x}\right)}.\]
Now, use the properties of logarithms to write
\[ \begin{align} \ln{L} &= \lim_{x \to \infty} \left( \frac{1}{x} \ln{x}\right) \\ &= \lim_{x \to \infty} \frac{\ln{x}}{x}\end{align}.\]
The above limit is now an indeterminate form of \(\infty/\infty\), so you can use L'Hôpital's rule, obtaining
\[ \begin{align} \ln{L} &= \lim_{x \to \infty} \frac{\frac{1}{x}}{1} \\ &=\frac{0}{1} \\&= 0.\end{align}\]
Finally, undo the natural logarithm by taking the exponential, which means that
\[ \begin{align} L &= e^0 \\ &= 1. \end{align} \]
Indeterminate Forms - Key takeaways
- An indeterminate form is an expression of two functions whose limit cannot be evaluated by direct substitution.
- The most common indeterminate forms are \(0/0\) and \( \pm\infty/\pm\infty\).
- The other indeterminate forms refer to the expressions \(0 \cdot \infty\), \(0^0\), \( \infty^0\), \(1^\infty\), and \(\infty-\infty\).
- A limit resulting in an indeterminate form that involves quotients can be evaluated using L'Hôpital's Rule.
- By algebraic means, it is possible to transform limits that involve the other indeterminate forms into expressions that involve quotients, so you can use L'Hôpital's rule to evaluate them.
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