Indeterminate Form of Type [latex]0 \cdot \infty[/latex]

Suppose we want to evaluate [latex]\underset{x\to a}{\lim}(f(x) \cdot g(x))[/latex], where [latex]f(x)\to 0[/latex] and [latex]g(x)\to \infty[/latex] (or [latex]−\infty[/latex]) as [latex]x\to a[/latex]. 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 [latex]0 \cdot \infty[/latex] to denote the form that arises in this situation. The expression [latex]0 \cdot \infty[/latex] is considered indeterminate because we cannot determine without further analysis the exact behavior of the product [latex]f(x)g(x)[/latex] as [latex]x\to {a}[/latex]. For example, let [latex]n[/latex] be a positive integer and consider

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

Example: Indeterminate Form of Type [latex]0·\infty[/latex]

Evaluate [latex]\underset{x\to 0^+}{\lim}x \ln x[/latex]

Show Solution

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

[latex]x \ln x=\frac{\ln x}{1/x}[/latex],

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

[latex]\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[/latex].

We conclude that

[latex]\underset{x\to 0^+}{\lim}x \ln x=0[/latex].

Figure 2. Finding the limit at [latex]x=0[/latex] of the function [latex]f(x)=x \ln x[/latex].

Indeterminate Form of Type [latex]\infty -\infty[/latex]

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

On the other hand, if [latex]n=2[/latex], then

However, if [latex]n=1[/latex], then

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

Watch the following video to see the worked solution to Example: Indeterminate Form of Type [latex]\infty -\infty[/latex].

Another type of indeterminate form that arises when evaluating limits involves exponents. The expressions [latex]0^0[/latex], [latex]\infty^0[/latex], and [latex]1^{\infty}[/latex] 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 [latex]\underset{x\to a}{\lim}f(x)^{g(x)}[/latex] and we arrive at the indeterminate form [latex]\infty^0[/latex]. (The indeterminate forms [latex]0^0[/latex] and [latex]1^{\infty}[/latex] can be handled similarly.) We proceed as follows. Let

Then,

Therefore,

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

Since the natural logarithm function is continuous, we conclude that

which gives us

Watch the following video to see the worked solution to Example: Indeterminate Form of Type [latex]\infty^0[/latex].

Watch the following video to see the worked solution to Example: Indeterminate Form of Type [latex]0^0[/latex].