L’Hôpital’s Rule: Learn It 2

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L’Hôpital’s Rule: Learn It 2

L’Hôpital’s Rule

Applying L’Hôpital’s Rule Cont.

L’Hôpital’s Rule ([latex]\infty / \infty[/latex] Case)

L’Hôpital’s Rule provides a method to resolve indeterminate forms in calculus, specifically those that result in [latex]\infty / \infty[/latex] when calculating limits.

L’Hôpital’s rule ([latex]\infty / \infty[/latex] case)

Suppose [latex]f[/latex] and [latex]g[/latex] are differentiable functions over an open interval containing [latex]a[/latex], except possibly at [latex]a[/latex]. Suppose [latex]\underset{x\to a}{\lim}f(x)=\infty[/latex] (or [latex]−\infty[/latex]) and [latex]\underset{x\to a}{\lim}g(x)=\infty[/latex] (or [latex]−\infty[/latex]). Then,

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

assuming the limit on the right exists or is [latex]\infty[/latex] or [latex]−\infty[/latex]. This result also holds if the limit is infinite, if [latex]a=\infty[/latex] or [latex]−\infty[/latex], or the limit is one-sided.

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].

To correctly apply L’Hôpital’s Rule to a quotient [latex]\frac{f(x)}{g(x)}[/latex], it is essential that the original limit of the quotient is an indeterminate form, either [latex]0/0[/latex] or [latex]\infty / \infty[/latex]. This is crucial because applying the rule outside these conditions does not yield valid results.

While L’Hôpital’s Rule is an invaluable tool for calculus, its application should be carefully considered. Not all limits of the form [latex]\infty / \infty[/latex] are suitable for L’Hôpital’s Rule without additional analysis or transformation of the function.

Consider the following non-applicable example to better understand the limitations:

Consider [latex]\underset{x\to 1}{\lim}\dfrac{x^2+5}{3x+4}[/latex]. Show that the limit cannot be evaluated by applying L’Hôpital’s rule.

Because the limits of the numerator and denominator are not both zero and are not both infinite, we cannot apply L’Hôpital’s rule. If we try to do so, we get

[latex]\frac{d}{dx}(x^2+5)=2x[/latex]

and,

[latex]\frac{d}{dx}(3x+4)=3[/latex]

At which point we would conclude erroneously that

[latex]\underset{x\to 1}{\lim}\frac{x^2+5}{3x+4}=\underset{x\to 1}{\lim}\frac{2x}{3}=\frac{2}{3}[/latex].

However, since [latex]\underset{x\to 1}{\lim}(x^2+5)=6[/latex] and [latex]\underset{x\to 1}{\lim}(3x+4)=7[/latex], we actually have:

[latex]\underset{x\to 1}{\lim}\frac{x^2+5}{3x+4}=\frac{6}{7}[/latex]

We can conclude that

[latex]\underset{x\to 1}{\lim}\frac{x^2+5}{3x+4}\ne \underset{x\to 1}{\lim}\frac{\frac{d}{dx}(x^2+5)}{\frac{d}{dx}(3x+4)}[/latex].[/hidden-answer]

Explain why we cannot apply L’Hôpital’s rule to evaluate [latex]\underset{x\to 0^+}{\lim}\dfrac{\cos x}{x}[/latex]. Evaluate [latex]\underset{x\to 0^+}{\lim}\dfrac{\cos x}{x}[/latex] by other means.