Double Integration and Applications: Background You’ll Need 3

Giselle Miller

Double Integration and Applications: Background You’ll Need 3

Recognize indeterminate forms and apply L’Hôpital’s rule for quotients, products, subtractions, and powers

L’Hôpital’s Rule

L’Hôpital’s Rule uses derivatives to evaluate limits that produce indeterminate forms [latex]\frac{0}{0}[/latex] or [latex]\frac{\infty}{\infty}[/latex].

L’Hôpital’s rule

Suppose [latex]f[/latex] and [latex]g[/latex] are differentiable near [latex]a[/latex]. If [latex]\underset{x\to a}{\lim}\dfrac{f(x)}{g(x)}[/latex] produces [latex]\frac{0}{0}[/latex] or [latex]\frac{\infty}{\infty}[/latex], then:

[latex]\underset{x\to a}{\lim}\dfrac{f(x)}{g(x)}=\underset{x\to a}{\lim}\dfrac{f'(x)}{g'(x)}[/latex]

provided the limit on the right exists or equals [latex]\pm\infty[/latex].

This also applies to one-sided limits and when [latex]a = \infty[/latex] or [latex]-\infty[/latex].

Critical Points:

Always verify the limit produces [latex]\frac{0}{0}[/latex] or [latex]\frac{\infty}{\infty}[/latex] before applying L’Hôpital’s Rule
Differentiate numerator and denominator **separately** (this is not the quotient rule)
You may need to apply L’Hôpital’s Rule multiple times
If the limit doesn’t produce an indeterminate form, do not use L’Hôpital’s Rule

Evaluate each of the following limits by applying L’Hôpital’s rule.

[latex]\underset{x\to 0}{\lim}\dfrac{1- \cos x}{x}[/latex]
[latex]\underset{x\to 1}{\lim}\dfrac{\sin (\pi x)}{\ln x}[/latex]
[latex]\underset{x\to \infty }{\lim}\dfrac{e^{\frac{1}{x}}-1}{\frac{1}{x}}[/latex]
[latex]\underset{x\to 0}{\lim}\dfrac{\sin x-x}{x^2}[/latex]

Since the numerator [latex]1- \cos x\to 0[/latex] and the denominator [latex]x\to 0[/latex], we can apply L’Hôpital’s rule to evaluate this limit. We have
[latex]\begin{array}{ll} \underset{x\to 0}{\lim}\frac{1- \cos x}{x} & =\underset{x\to 0}{\lim}\frac{\frac{d}{dx}(1- \cos x)}{\frac{d}{dx}(x)} \\ & =\underset{x\to 0}{\lim}\frac{\sin x}{1} \\ & =\frac{\underset{x\to 0}{\lim}(\sin x)}{\underset{x\to 0}{\lim}(1)} \\ & =\frac{0}{1}=0 \end{array}[/latex]
As [latex]x\to 1[/latex], the numerator [latex]\sin (\pi x)\to 0[/latex] and the denominator [latex]\ln x \to 0[/latex]. Therefore, we can apply L’Hôpital’s rule. We obtain
[latex]\begin{array}{ll} \underset{x\to 1}{\lim}\frac{\sin (\pi x)}{\ln x} & =\underset{x\to 1}{\lim}\frac{\pi \cos (\pi x)}{1/x} \\ & =\underset{x\to 1}{\lim}(\pi x) \cos (\pi x) \\ & =(\pi \cdot 1)(-1)=−\pi \end{array}[/latex]
As [latex]x\to \infty[/latex], the numerator [latex]e^{1/x}-1\to 0[/latex] and the denominator [latex](\frac{1}{x})\to 0[/latex]. Therefore, we can apply L’Hôpital’s rule. We obtain
[latex]\underset{x\to \infty }{\lim}\frac{e^{1/x}-1}{\frac{1}{x}}=\underset{x\to \infty }{\lim}\frac{e^{1/x}(\frac{-1}{x^2})}{(\frac{-1}{x^2})}=\underset{x\to \infty}{\lim} e^{1/x}=e^0=1[/latex]
As [latex]x\to 0[/latex], both the numerator and denominator approach zero. Therefore, we can apply L’Hôpital’s rule. We obtain
[latex]\underset{x\to 0}{\lim}\frac{\sin x-x}{x^2}=\underset{x\to 0}{\lim}\frac{\cos x-1}{2x}[/latex].

Since the numerator and denominator of this new quotient both approach zero as [latex]x\to 0[/latex], we apply L’Hôpital’s rule again. In doing so, we see that

[latex]\underset{x\to 0}{\lim}\frac{\cos x-1}{2x}=\underset{x\to 0}{\lim}\frac{−\sin x}{2}=0[/latex].

Therefore, we conclude that

[latex]\underset{x\to 0}{\lim}\frac{\sin x-x}{x^2}=0[/latex].

Watch the following video to see the worked solution to this example.

For closed captioning, open the video on its original page by clicking the Youtube logo in the lower right-hand corner of the video display. In YouTube, the video will begin at the same starting point as this clip, but will continue playing until the very end.

You can view the transcript for this segmented clip of “4.8 L’Hopital’s Rule” here (opens in new window).

Evaluate each of the following limits by applying L’Hôpital’s rule.

[latex]\underset{x\to \infty }{\lim}\dfrac{3x+5}{2x+1}[/latex]
[latex]\underset{x\to 0^+}{\lim}\dfrac{\ln x}{\cot x}[/latex]

Since [latex]3x+5[/latex] and [latex]2x+1[/latex] are first-degree polynomials with positive leading coefficients, [latex]\underset{x\to \infty }{\lim}(3x+5)=\infty[/latex] and [latex]\underset{x\to \infty }{\lim}(2x+1)=\infty[/latex]. Therefore, we apply L’Hôpital’s rule and obtain
[latex]\underset{x\to \infty}{\lim}\dfrac{3x+5}{2x+\frac{1}{x}}=\underset{x\to \infty }{\lim}\dfrac{3}{2}=\dfrac{3}{2}[/latex].

Note that this limit can also be calculated without invoking L’Hôpital’s rule. Earlier in the module we showed how to evaluate such a limit by dividing the numerator and denominator by the highest power of [latex]x[/latex] in the denominator. In doing so, we saw that

[latex]\underset{x\to \infty }{\lim}\dfrac{3x+5}{2x+1}=\underset{x\to \infty }{\lim}\dfrac{3+\frac{5}{x}}{2+\frac{1}{x}}=\dfrac{3}{2}[/latex].

L’Hôpital’s rule provides us with an alternative means of evaluating this type of limit.
Here, [latex]\underset{x\to 0^+}{\lim} \ln x=−\infty[/latex] and [latex]\underset{x\to 0^+}{\lim} \cot x=\infty[/latex]. Therefore, we can apply L’Hôpital’s rule and obtain
[latex]\underset{x\to 0^+}{\lim}\dfrac{\ln x}{\cot x}=\underset{x\to 0^+}{\lim}\dfrac{\frac{1}{x}}{−\csc^2 x}=\underset{x\to 0^+}{\lim}\dfrac{1}{−x \csc^2 x}[/latex].

Now as [latex]x\to 0^+[/latex], [latex]\csc^2 x\to \infty[/latex]. Therefore, the first term in the denominator is approaching zero and the second term is getting really large. In such a case, anything can happen with the product. Therefore, we cannot make any conclusion yet. To evaluate the limit, we use the definition of [latex]\csc x[/latex] to write

[latex]\underset{x\to 0^+}{\lim}\dfrac{1}{−x \csc^2 x}=\underset{x\to 0^+}{\lim}\dfrac{\sin^2 x}{−x}[/latex].

Now [latex]\underset{x\to 0^+}{\lim} \sin^2 x=0[/latex] and [latex]\underset{x\to 0^+}{\lim} x=0[/latex], so we apply L’Hôpital’s rule again. We find

[latex]\underset{x\to 0^+}{\lim}\dfrac{\sin^2 x}{−x}=\underset{x\to 0^+}{\lim}\dfrac{2 \sin x \cos x}{-1}=\dfrac{0}{-1}=0[/latex].

We conclude that

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

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

When [latex]\underset{x\to a}{\lim}f(x) = 0[/latex] and [latex]\underset{x\to a}{\lim}g(x) = \infty[/latex], the product [latex]f(x)g(x)[/latex] produces the indeterminate form [latex]0 \cdot \infty[/latex]. To apply L’Hôpital’s Rule, rewrite the product as a quotient:

[latex]f(x) \cdot g(x) = \dfrac{f(x)}{\frac{1}{g(x)}}[/latex] or [latex]f(x) \cdot g(x) = \dfrac{g(x)}{\frac{1}{f(x)}}[/latex]

Choose whichever form gives [latex]\frac{0}{0}[/latex] or [latex]\frac{\infty}{\infty}[/latex].

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

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

When both [latex]f(x) \to \infty[/latex] and [latex]g(x) \to \infty[/latex] as [latex]x \to a[/latex], the difference [latex]f(x) - g(x)[/latex] produces [latex]\infty - \infty[/latex]. To use L’Hôpital’s Rule, combine the terms into a single quotient, typically by finding a common denominator.

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

Indeterminate Forms Involving Exponents

The forms [latex]0^0[/latex], [latex]\infty^0[/latex], and [latex]1^\infty[/latex] are indeterminate when they arise in limits. To handle limits of the form [latex]\underset{x\to a}{\lim}f(x)^{g(x)}[/latex], use logarithms.

Strategy for Exponential Indeterminate Forms

Let [latex]y = f(x)^{g(x)}[/latex]
Take the natural logarithm: [latex]\ln y = g(x) \ln(f(x))[/latex]
Evaluate [latex]L = \underset{x\to a}{\lim}g(x)\ln(f(x))[/latex] using techniques for [latex]0 \cdot \infty[/latex]
Then [latex]\underset{x\to a}{\lim}f(x)^{g(x)} = e^L[/latex]

Evaluate [latex]\underset{x\to \infty }{\lim} x^{\frac{1}{x}}[/latex]