{"id":321,"date":"2023-09-20T22:49:04","date_gmt":"2023-09-20T22:49:04","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/calculus1\/chapter\/other-indeterminate-forms\/"},"modified":"2025-04-17T13:30:06","modified_gmt":"2025-04-17T13:30:06","slug":"lhopitals-rule-learn-it-3","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/calculus1\/chapter\/lhopitals-rule-learn-it-3\/","title":{"raw":"L\u2019H\u00f4pital\u2019s Rule: Learn It 3","rendered":"L\u2019H\u00f4pital\u2019s Rule: Learn It 3"},"content":{"raw":"

Other Indeterminate Forms<\/h2>\r\n

L\u2019H\u00f4pital\u2019s rule is very useful for evaluating limits involving the indeterminate forms [latex]\\frac{0}{0}[\/latex] and [latex]\\frac{\\infty}{\\infty}[\/latex]. However, we can also use L\u2019H\u00f4pital\u2019s rule to help evaluate limits involving other indeterminate forms that arise when evaluating limits.<\/p>\r\n

The expressions [latex]0 \\cdot \\infty[\/latex], [latex]\\infty - \\infty[\/latex], [latex]1^{\\infty}[\/latex], [latex]\\infty^0[\/latex], and [latex]0^0[\/latex] are all considered indeterminate forms. These expressions are not real numbers. Rather, they represent forms that arise when trying to evaluate certain limits.<\/p>\r\n

Next we realize why these are indeterminate forms and then understand how to use L\u2019H\u00f4pital\u2019s 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 [latex]\\frac{0}{0}[\/latex] or [latex]\\frac{\\infty}{\\infty}[\/latex].<\/p>\r\n

Indeterminate Form of Type [latex]0 \\cdot \\infty[\/latex]<\/h3>\r\n

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]\u2212\\infty[\/latex]) as [latex]x\\to a[\/latex].<\/p>\r\n

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.<\/p>\r\n

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<\/p>\r\n

[latex]f(x)=\\dfrac{1}{(x^n+1)}[\/latex] and [latex]g(x)=3x^2[\/latex].<\/div>\r\n

As [latex]x\\to \\infty[\/latex], [latex]f(x)\\to 0[\/latex] and [latex]g(x)\\to \\infty[\/latex].<\/p>\r\n

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].<\/p>\r\n

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\u2019H\u00f4pital\u2019s rule.<\/p>\r\n

\r\n

Evaluate [latex]\\underset{x\\to 0^+}{\\lim}x \\ln x[\/latex].<\/p>\r\n

First, rewrite the function [latex]x \\ln x[\/latex] as a quotient to apply L\u2019H\u00f4pital\u2019s rule. If we write<\/p>\r\n

[latex]x \\ln x=\\frac{\\ln x}{1\/x}[\/latex],<\/div>\r\n

we see that [latex]\\ln x\\to \u2212\\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\u2019H\u00f4pital\u2019s rule and obtain<\/p>\r\n

[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}(\u2212x)=0[\/latex].<\/div>\r\n

We conclude that<\/p>\r\n

[latex]\\underset{x\\to 0^+}{\\lim}x \\ln x=0[\/latex].<\/div>\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"358\"]\"The Figure 2. Finding the limit at [latex]x=0[\/latex] of the function [latex]f(x)=x \\ln x[\/latex].[\/caption]\r\n<\/section>\r\n
\r\n

Evaluate [latex]\\underset{x\\to 0}{\\lim}x \\cot x[\/latex].<\/p>\r\n

[reveal-answer q=\"404616\"]Hint[\/reveal-answer]
\r\n[hidden-answer a=\"404616\"]<\/p>\r\n

Write [latex]x \\cot x=\\frac{x \\cos x}{\\sin x}[\/latex]<\/p>\r\n

[\/hidden-answer]<\/p>\r\n

[reveal-answer q=\"fs-id1165043286669\"]Show Solution[\/reveal-answer]
\r\n[hidden-answer a=\"fs-id1165043286669\"]<\/p>\r\n

[latex]1[\/latex]<\/p>\r\n

[\/hidden-answer]<\/p>\r\n<\/section>\r\n

Indeterminate Form of Type [latex]\\infty -\\infty[\/latex]<\/h3>\r\n

Another type of indeterminate form is [latex]\\infty -\\infty[\/latex]. Consider the following example:<\/p>\r\n

\r\n

Let [latex]n[\/latex] be a positive integer and let [latex]f(x)=3x^n[\/latex] and [latex]g(x)=3x^2+5[\/latex].<\/p>\r\n

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].<\/p>\r\n

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.<\/p>\r\n<\/section>\r\n

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.<\/p>\r\n

\r\n

Suppose the exponent [latex]n[\/latex] in the function [latex]f(x)=3x^n[\/latex] is [latex]n=3[\/latex], then<\/p>\r\n

[latex]\\underset{x\\to \\infty }{\\lim}(f(x)-g(x))=\\underset{x\\to \\infty }{\\lim}(3x^3-3x^2-5)=\\infty[\/latex].<\/div>\r\n

On the other hand, if [latex]n=2[\/latex], then<\/p>\r\n

[latex]\\underset{x\\to \\infty }{\\lim}(f(x)-g(x))=\\underset{x\\to \\infty }{\\lim}(3x^2-3x^2-5)=-5[\/latex].<\/div>\r\n

However, if [latex]n=1[\/latex], then<\/p>\r\n

[latex]\\underset{x\\to \\infty }{\\lim}(f(x)-g(x))=\\underset{x\\to \\infty }{\\lim}(3x-3x^2-5)=\u2212\\infty[\/latex].<\/div>\r\n

Therefore, the limit cannot be determined by considering only [latex]\\infty -\\infty[\/latex].\u00a0<\/p>\r\n<\/section>\r\n

Next we see how to rewrite an expression involving the indeterminate form [latex]\\infty -\\infty [\/latex] as a fraction to apply L\u2019H\u00f4pital\u2019s rule.<\/div>\r\n
\r\n

Evaluate [latex]\\underset{x\\to 0^+}{\\lim}\\left(\\dfrac{1}{x^2}-\\dfrac{1}{\\tan x}\\right)[\/latex].<\/p>\r\n

By combining the fractions, we can write the function as a quotient. Since the least common denominator is [latex]x^2 \\tan x[\/latex], we have<\/p>\r\n

[latex]\\frac{1}{x^2}-\\frac{1}{\\tan x}=\\frac{(\\tan x)-x^2}{x^2 \\tan x}[\/latex]<\/div>\r\n

As [latex]x\\to 0^+[\/latex], the numerator [latex]\\tan x-x^2 \\to 0[\/latex] and the denominator [latex]x^2 \\tan x \\to 0[\/latex]. Therefore, we can apply L\u2019H\u00f4pital\u2019s rule. Taking the derivatives of the numerator and the denominator, we have<\/p>\r\n

[latex]\\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}[\/latex]<\/div>\r\n

As [latex]x\\to 0^+[\/latex], [latex](\\sec^2 x)-2x \\to 1[\/latex] and [latex]x^2 \\sec^2 x+2x \\tan x \\to 0[\/latex]. Since the denominator is positive as [latex]x[\/latex] approaches zero from the right, we conclude that<\/p>\r\n

[latex]\\underset{x\\to 0^+}{\\lim}\\frac{(\\sec^2 x)-2x}{x^2 \\sec^2 x+2x \\tan x}=\\infty[\/latex]<\/div>\r\n

Therefore,<\/p>\r\n

[latex]\\underset{x\\to 0^+}{\\lim}(\\frac{1}{x^2}-\\frac{1}{ tan x})=\\infty[\/latex]\u00a0<\/span><\/div>\r\n<\/section>\r\n
\r\n

Watch the following video to see the worked solution to this example.<\/p>\r\n