Other Indeterminate Forms

Learning Outcomes

Identify indeterminate forms produced by quotients, products, subtractions, and powers, and apply L’Hôpital’s rule in each case

L’Hôpital’s rule is very useful for evaluating limits involving the indeterminate forms $\frac{0}{0}$ and $\frac{\infty}{\infty}$ . However, we can also use L’Hôpital’s rule to help evaluate limits involving other indeterminate forms that arise when evaluating limits. The expressions $0 \cdot \infty$ , $\infty - \infty$ , $1^{\infty}$ , $\infty^0$ , and $0^0$ are all considered indeterminate forms. These expressions are not real numbers. Rather, they represent forms that arise when trying to evaluate certain limits. Next we realize why these are indeterminate forms and then understand how to use L’Hôpital’s rule in these cases. The key idea is that we must rewrite the indeterminate forms in such a way that we arrive at the indeterminate form $\frac{0}{0}$ or $\frac{\infty}{\infty}$ .

Indeterminate Form of Type $0 \cdot \infty$

Suppose we want to evaluate $\underset{x\to a}{\lim}(f(x) \cdot g(x))$ , where $f(x)\to 0$ and $g(x)\to \infty$ (or $−\infty$ ) as $x\to a$ . Since one term in the product is approaching zero but the other term is becoming arbitrarily large (in magnitude), anything can happen to the product. We use the notation $0 \cdot \infty$ to denote the form that arises in this situation. The expression $0 \cdot \infty$ is considered indeterminate because we cannot determine without further analysis the exact behavior of the product $f(x)g(x)$ as $x\to {a}$ . For example, let $n$ be a positive integer and consider

f (x) = \frac{1}{(x^{n} + 1)}

$f(x)=\dfrac{1}{(x^n+1)}$ and

g (x) = 3 x^{2}

$g(x)=3x^2$ .

As $x\to \infty$ , $f(x)\to 0$ and $g(x)\to \infty$ . However, the limit as $x\to \infty$ of $f(x)g(x)=\frac{3x^2}{(x^n+1)}$ varies, depending on $n$ . If $n=2$ , then $\underset{x\to \infty }{\lim}f(x)g(x)=3$ . If $n=1$ , then $\underset{x\to \infty }{\lim}f(x)g(x)=\infty$ . If $n=3$ , then $\underset{x\to \infty }{\lim}f(x)g(x)=0$ . Here we consider another limit involving the indeterminate form $0 \cdot \infty$ and show how to rewrite the function as a quotient to use L’Hôpital’s rule.

Example: Indeterminate Form of Type $0·\infty$

Evaluate $\underset{x\to 0^+}{\lim}x \ln x$

Show Solution

First, rewrite the function $x \ln x$ as a quotient to apply L’Hôpital’s rule. If we write

x \ln x = \frac{\ln x}{1 / x}

$x \ln x=\frac{\ln x}{1/x}$ ,

we see that $\ln x\to −\infty$ as $x\to 0^+$ and $\frac{1}{x}\to \infty$ as $x\to 0^+$ . Therefore, we can apply L’Hôpital’s rule and obtain

lim_{x \to 0^{+}} \frac{\ln x}{1 / x} = lim_{x \to 0^{+}} \frac{\frac{d}{d x} (\ln x)}{\frac{d}{d x} (1 / x)} = lim_{x \to 0^{+}} \frac{1 / x}{- 1 / x^{2}} = lim_{x \to 0^{+}} (- x) = 0

$\underset{x\to 0^+}{\lim}\frac{\ln x}{1/x}=\underset{x\to 0^+}{\lim}\frac{\frac{d}{dx}(\ln x)}{\frac{d}{dx}(1/x)}=\underset{x\to 0^+}{\lim}\frac{1/x}{-1/x^2}=\underset{x\to 0^+}{\lim}(−x)=0$ .

We conclude that

lim_{x \to 0^{+}} x \ln x = 0

$\underset{x\to 0^+}{\lim}x \ln x=0$ .

The function y = x ln(x) is graphed for values x ≥ 0. At x = 0, the value of the function is 0.

Figure 2. Finding the limit at $x=0$ of the function $f(x)=x \ln x$ .

Try It

Evaluate $\underset{x\to 0}{\lim}x \cot x$

Hint

Write $x \cot x=\frac{x \cos x}{\sin x}$

Show Solution

Indeterminate Form of Type $\infty -\infty$

Another type of indeterminate form is $\infty -\infty$ . Consider the following example. Let $n$ be a positive integer and let $f(x)=3x^n$ and $g(x)=3x^2+5$ . As $x\to \infty$ , $f(x)\to \infty$ and $g(x)\to \infty$ . We are interested in $\underset{x\to \infty}{\lim}(f(x)-g(x))$ . Depending on whether $f(x)$ grows faster, $g(x)$ grows faster, or they grow at the same rate, as we see next, anything can happen in this limit. Since $f(x)\to \infty$ and $g(x)\to \infty$ , we write $\infty -\infty$ to denote the form of this limit. As with our other indeterminate forms, $\infty -\infty$ has no meaning on its own and we must do more analysis to determine the value of the limit. For example, suppose the exponent $n$ in the function $f(x)=3x^n$ is $n=3$ , then

lim_{x \to \infty} (f (x) - g (x)) = lim_{x \to \infty} (3 x^{3} - 3 x^{2} - 5) = \infty

$\underset{x\to \infty }{\lim}(f(x)-g(x))=\underset{x\to \infty }{\lim}(3x^3-3x^2-5)=\infty$ .

On the other hand, if $n=2$ , then

lim_{x \to \infty} (f (x) - g (x)) = lim_{x \to \infty} (3 x^{2} - 3 x^{2} - 5) = - 5

$\underset{x\to \infty }{\lim}(f(x)-g(x))=\underset{x\to \infty }{\lim}(3x^2-3x^2-5)=-5$ .

However, if $n=1$ , then

lim_{x \to \infty} (f (x) - g (x)) = lim_{x \to \infty} (3 x - 3 x^{2} - 5) = - \infty

$\underset{x\to \infty }{\lim}(f(x)-g(x))=\underset{x\to \infty }{\lim}(3x-3x^2-5)=−\infty$ .

Therefore, the limit cannot be determined by considering only $\infty -\infty$ . Next we see how to rewrite an expression involving the indeterminate form $\infty -\infty$ as a fraction to apply L’Hôpital’s rule.

Example: Indeterminate Form of Type $\infty -\infty$

Evaluate $\underset{x\to 0^+}{\lim}\left(\dfrac{1}{x^2}-\dfrac{1}{\tan x}\right)$ .

Show Solution

By combining the fractions, we can write the function as a quotient. Since the least common denominator is $x^2 \tan x$ , we have

\frac{1}{x^{2}} - \frac{1}{\tan x} = \frac{(\tan x) - x^{2}}{x^{2} \tan x}

$\frac{1}{x^2}-\frac{1}{\tan x}=\frac{(\tan x)-x^2}{x^2 \tan x}$

As $x\to 0^+$ , the numerator $\tan x-x^2 \to 0$ and the denominator $x^2 \tan x \to 0$ . Therefore, we can apply L’Hôpital’s rule. Taking the derivatives of the numerator and the denominator, we have

lim_{x \to 0^{+}} \frac{(\tan x) - x^{2}}{x^{2} \tan x} = lim_{x \to 0^{+}} \frac{(\sec^{2} x) - 2 x}{x^{2} \sec^{2} x + 2 x \tan x}

$\underset{x\to 0^+}{\lim}\frac{(\tan x)-x^2}{x^2 \tan x}=\underset{x\to 0^+}{\lim}\frac{(\sec^2 x)-2x}{x^2 \sec^2 x+2x \tan x}$

As $x\to 0^+$ , $(\sec^2 x)-2x \to 1$ and $x^2 \sec^2 x+2x \tan x \to 0$ . Since the denominator is positive as $x$ approaches zero from the right, we conclude that

lim_{x \to 0^{+}} \frac{(\sec^{2} x) - 2 x}{x^{2} \sec^{2} x + 2 x \tan x} = \infty

$\underset{x\to 0^+}{\lim}\frac{(\sec^2 x)-2x}{x^2 \sec^2 x+2x \tan x}=\infty$

Therefore,

lim_{x \to 0^{+}} (\frac{1}{x^{2}} - \frac{1}{t a n x}) = \infty

$\underset{x\to 0^+}{\lim}(\frac{1}{x^2}-\frac{1}{ tan x})=\infty$

Watch the following video to see the worked solution to Example: Indeterminate Form of Type $\infty -\infty$ .

Closed Captioning and Transcript Information for Video

Try It

Evaluate $\underset{x\to 0^+}{\lim}\left(\dfrac{1}{x}-\dfrac{1}{\sin x}\right)$ .

Hint

Show Solution

Another type of indeterminate form that arises when evaluating limits involves exponents. The expressions $0^0$ , $\infty^0$ , and $1^{\infty}$ are all indeterminate forms. On their own, these expressions are meaningless because we cannot actually evaluate these expressions as we would evaluate an expression involving real numbers. Rather, these expressions represent forms that arise when finding limits. Now we examine how L’Hôpital’s rule can be used to evaluate limits involving these indeterminate forms.

Since L’Hôpital’s rule applies to quotients, we use the natural logarithm function and its properties to reduce a problem evaluating a limit involving exponents to a related problem involving a limit of a quotient. For example, suppose we want to evaluate $\underset{x\to a}{\lim}f(x)^{g(x)}$ and we arrive at the indeterminate form $\infty^0$ . (The indeterminate forms $0^0$ and $1^{\infty}$ can be handled similarly.) We proceed as follows. Let

y = f (x)^{g (x)}

$y=f(x)^{g(x)}$

Then,

\ln y = \ln (f (x)^{g (x)}) = g (x) \ln (f (x))

$\ln y=\ln (f(x)^{g(x)})=g(x) \ln (f(x))$

Therefore,

lim_{x \to a} [\ln y] = lim_{x \to a} [g (x) \ln (f (x))]

$\underset{x\to a}{\lim}[\ln y]=\underset{x\to a}{\lim}[g(x) \ln (f(x))]$

Since $\underset{x\to a}{\lim}f(x)=\infty$ , we know that $\underset{x\to a}{\lim}\ln (f(x))=\infty$ . Therefore, $\underset{x\to a}{\lim}g(x) \ln (f(x))$ is of the indeterminate form $0 \cdot \infty$ , and we can use the techniques discussed earlier to rewrite the expression $g(x) \ln (f(x))$ in a form so that we can apply L’Hôpital’s rule. Suppose $\underset{x\to a}{\lim}g(x) \ln (f(x))=L$ , where $L$ may be $\infty$ or $−\infty$ . Then

lim_{x \to a} \ln y = L

$\underset{x\to a}{\lim}\ln y=L$

Since the natural logarithm function is continuous, we conclude that

\ln (lim_{x \to a} y) = L

$\ln (\underset{x\to a}{\lim} y)=L$

which gives us

lim_{x \to a} y = lim_{x \to a} f (x)^{g (x)} = e^{L}

$\underset{x\to a}{\lim} y=\underset{x\to a}{\lim}f(x)^{g(x)}=e^L$

Example: Indeterminate Form of Type $\infty^0$

Evaluate $\underset{x\to \infty }{\lim} x^{\frac{1}{x}}$

Show Solution

Let $y=x^{1/x}$ . Then,

\ln (x^{1 / x}) = \frac{1}{x} \ln x = \frac{\ln x}{x}

$\ln (x^{1/x})=\frac{1}{x} \ln x=\frac{\ln x}{x}$

We need to evaluate $\underset{x\to \infty }{\lim}\frac{\ln x}{x}$ . Applying L’Hôpital’s rule, we obtain

lim_{x \to \infty} \ln y = lim_{x \to \infty} \frac{\ln x}{x} = lim_{x \to \infty} \frac{1 / x}{1} = 0

$\underset{x\to \infty }{\lim} \ln y=\underset{x\to \infty}{\lim}\frac{\ln x}{x}=\underset{x\to \infty}{\lim}\frac{1/x}{1}=0$

Therefore, $\underset{x\to \infty }{\lim}\ln y=0$ . Since the natural logarithm function is continuous, we conclude that

\ln (lim_{x \to \infty} y) = 0

$\ln (\underset{x\to \infty}{\lim} y)=0$ ,

which leads to

lim_{x \to \infty} y = lim_{x \to \infty} \frac{\ln x}{x} = e^{0} = 1

$\underset{x\to \infty }{\lim} y=\underset{x\to \infty}{\lim}\frac{\ln x}{x}=e^0=1$

Hence,

lim_{x \to \infty} x^{1 / x} = 1

$\underset{x\to \infty}{\lim} x^{1/x}=1$

Watch the following video to see the worked solution to Example: Indeterminate Form of Type $\infty^0$ .

Closed Captioning and Transcript Information for Video

Try It

Evaluate $\underset{x\to \infty}{\lim} x^{\frac{1}{\ln x}}$

Hint

Let $y=x^{1/ \ln x}$ and apply the natural logarithm to both sides of the equation.

Show Solution

$e$

Example: Indeterminate Form of Type $0^0$

Evaluate $\underset{x\to 0^+}{\lim} x^{\sin x}$

Show Solution

Let

y = x^{\sin x}

$y=x^{\sin x}$

Therefore,

\ln y = \ln (x^{\sin x}) = \sin x \ln x

$\ln y=\ln (x^{\sin x})= \sin x \ln x$

We now evaluate $\underset{x\to 0^+}{\lim} \sin x \ln x$ . Since $\underset{x\to 0^+}{\lim} \sin x=0$ and $\underset{x\to 0^+}{\lim} \ln x=−\infty$ , we have the indeterminate form $0 \cdot \infty$ . To apply L’Hôpital’s rule, we need to rewrite $\sin x \ln x$ as a fraction. We could write

\sin x \ln x = \frac{\sin x}{1 / \ln x}

$\sin x \ln x=\frac{\sin x}{1/ \ln x}$

\sin x \ln x = \frac{\ln x}{1 / \sin x} = \frac{\ln x}{\csc x}

$\sin x \ln x=\frac{\ln x}{1/ \sin x}=\frac{\ln x}{\csc x}$

Let’s consider the first option. In this case, applying L’Hôpital’s rule, we would obtain

lim_{x \to 0^{+}} \sin x \ln x = lim_{x \to 0^{+}} \frac{\sin x}{1 / \ln x} = lim_{x \to 0^{+}} \frac{\cos x}{- 1 / (x (\ln x)^{2})} = lim_{x \to 0^{+}} (- x (\ln x)^{2} \cos x)

$\underset{x\to 0^+}{\lim} \sin x \ln x=\underset{x\to 0^+}{\lim}\frac{\sin x}{1/ \ln x}=\underset{x\to 0^+}{\lim}\frac{\cos x}{-1/(x(\ln x)^2)}=\underset{x\to 0^+}{\lim}(−x(\ln x)^2 \cos x)$

Unfortunately, we not only have another expression involving the indeterminate form $0 \cdot \infty$ , but the new limit is even more complicated to evaluate than the one with which we started. Instead, we try the second option. By writing

\sin x \ln x = \frac{\ln x}{1 / \sin x} = \frac{\ln x}{\csc x}

$\sin x \ln x=\frac{\ln x}{1/ \sin x}=\frac{\ln x}{\csc x}$

and applying L’Hôpital’s rule, we obtain

lim_{x \to 0^{+}} \sin x \ln x = lim_{x \to 0^{+}} \frac{\ln x}{\csc x} = lim_{x \to 0^{+}} \frac{1 / x}{- \csc x \cot x} = lim_{x \to 0^{+}} \frac{- 1}{x \csc x \cot x}

$\underset{x\to 0^+}{\lim} \sin x \ln x=\underset{x\to 0^+}{\lim}\frac{\ln x}{\csc x}=\underset{x\to 0^+}{\lim}\frac{1/x}{− \csc x \cot x}=\underset{x\to 0^+}{\lim}\frac{-1}{x \csc x \cot x}$

Using the fact that $\csc x=\frac{1}{\sin x}$ and $\cot x=\frac{\cos x}{\sin x}$ , we can rewrite the expression on the right-hand side as

lim_{x \to 0^{+}} \frac{- \sin^{2} x}{x \cos x} = lim_{x \to 0^{+}} [\frac{\sin x}{x} \cdot (- \tan x)] = (lim_{x \to 0^{+}} \frac{\sin x}{x}) \cdot (lim_{x \to 0^{+}} (- \tan x)) = 1 \cdot 0 = 0

$\underset{x\to 0^+}{\lim}\frac{−\sin^2 x}{x \cos x}=\underset{x\to 0^+}{\lim}[\frac{\sin x}{x} \cdot (−\tan x)]=(\underset{x\to 0^+}{\lim}\frac{\sin x}{x}) \cdot (\underset{x\to 0^+}{\lim}(−\tan x))=1 \cdot 0=0$

We conclude that $\underset{x\to 0^+}{\lim} \ln y=0$ . Therefore, $\ln (\underset{x\to 0^+}{\lim} y)=0$ and we have

lim_{x \to 0^{+}} y = lim_{x \to 0^{+}} x^{\sin x} = e^{0} = 1

$\underset{x\to 0^+}{\lim} y=\underset{x\to 0^+}{\lim} x^{\sin x}=e^0=1$

Hence,

lim_{x \to 0^{+}} x^{\sin x} = 1

$\underset{x\to 0^+}{\lim} x^{\sin x}=1$

Watch the following video to see the worked solution to Example: Indeterminate Form of Type $0^0$ .

Closed Captioning and Transcript Information for Video

Try It

Evaluate $\underset{x\to 0^+}{\lim} x^x$

Hint

Let $y=x^x$ and take the natural logarithm of both sides of the equation.

Show Solution

Module 4: Applications of Derivatives