Applying L’Hôpital’s Rule

Learning Outcomes

  • Recognize when to apply L’Hôpital’s rule

Applying L’Hôpital’s Rule

L’Hôpital’s rule can be used to evaluate limits involving the quotient of two functions. Consider

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

 

If [latex]\underset{x\to a}{\lim}f(x)=L_1[/latex] and [latex]\underset{x\to a}{\lim}g(x)=L_2 \ne 0[/latex], then

[latex]\underset{x\to a}{\lim}\dfrac{f(x)}{g(x)}=\dfrac{L_1}{L_2}[/latex]

 

However, what happens if [latex]\underset{x\to a}{\lim}f(x)=0[/latex] and [latex]\underset{x\to a}{\lim}g(x)=0[/latex]? We call this one of the indeterminate forms, of type [latex]\frac{0}{0}[/latex]. This is considered an indeterminate form because we cannot determine the exact behavior of [latex]\frac{f(x)}{g(x)}[/latex] as [latex]x\to a[/latex] without further analysis. We have seen examples of this earlier in the text. For example, consider

[latex]\underset{x\to 2}{\lim}\dfrac{x^2-4}{x-2}[/latex] and [latex]\underset{x\to 0}{\lim}\dfrac{ \sin x}{x}[/latex]

 

For the first of these examples, we can evaluate the limit by factoring the numerator and writing

[latex]\underset{x\to 2}{\lim}\dfrac{x^2-4}{x-2}=\underset{x\to 2}{\lim}\dfrac{(x+2)(x-2)}{x-2}=\underset{x\to 2}{\lim}(x+2)=2+2=4[/latex]

For [latex]\underset{x\to 0}{\lim}\frac{\sin x}{x}[/latex] we were able to show, using a geometric argument, that

[latex]\underset{x\to 0}{\lim}\dfrac{\sin x}{x}=1[/latex]

 

Here we use a different technique for evaluating limits such as these. Not only does this technique provide an easier way to evaluate these limits, but also, and more important, it provides us with a way to evaluate many other limits that we could not calculate previously.

The idea behind L’Hôpital’s rule can be explained using local linear approximations. Consider two differentiable functions [latex]f[/latex] and [latex]g[/latex] such that [latex]\underset{x\to a}{\lim}f(x)=0=\underset{x\to a}{\lim}g(x)[/latex] and such that [latex]g^{\prime}(a)\ne 0[/latex] For [latex]x[/latex] near [latex]a[/latex], we can write

[latex]f(x)\approx f(a)+f^{\prime}(a)(x-a)[/latex]

 

and

[latex]g(x)\approx g(a)+g^{\prime}(a)(x-a)[/latex]

 

Therefore,

[latex]\dfrac{f(x)}{g(x)}\approx \dfrac{f(a)+f^{\prime}(a)(x-a)}{g(a)+g^{\prime}(a)(x-a)}[/latex]

 

Two functions y = f(x) and y = g(x) are drawn such that they cross at a point above x = a. The linear approximations of these two functions y = f(a) + f’(a)(x – a) and y = g(a) + g’(a)(x – a) are also drawn.

Figure 1. If [latex]\underset{x\to a}{\lim}f(x)=\underset{x\to a}{\lim}g(x)[/latex], then the ratio [latex]f(x)/g(x)[/latex] is approximately equal to the ratio of their linear approximations near [latex]a[/latex].

Since [latex]f[/latex] is differentiable at [latex]a[/latex], then [latex]f[/latex] is continuous at [latex]a[/latex], and therefore [latex]f(a)=\underset{x\to a}{\lim}f(x)=0[/latex]. Similarly, [latex]g(a)=\underset{x\to a}{\lim}g(x)=0[/latex]. If we also assume that [latex]f^{\prime}[/latex] and [latex]g^{\prime}[/latex] are continuous at [latex]x=a[/latex], then [latex]f^{\prime}(a)=\underset{x\to a}{\lim}f^{\prime}(x)[/latex] and [latex]g^{\prime}(a)=\underset{x\to a}{\lim}g^{\prime}(x)[/latex]. Using these ideas, we conclude that

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

Note that the assumption that [latex]f^{\prime}[/latex] and [latex]g^{\prime}[/latex] are continuous at [latex]a[/latex] and [latex]g^{\prime}(a)\ne 0[/latex] can be loosened. We state L’Hôpital’s rule formally for the indeterminate form [latex]\frac{0}{0}[/latex]. Also note that the notation [latex]\frac{0}{0}[/latex] does not mean we are actually dividing zero by zero. Rather, we are using the notation [latex]\frac{0}{0}[/latex] to represent a quotient of limits, each of which is zero.

L’Hôpital’s Rule (0/0 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]. If [latex]\underset{x\to a}{\lim}f(x)=0[/latex] and [latex]\underset{x\to a}{\lim}g(x)=0[/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 we are considering one-sided limits, or if [latex]a=\infty[/latex] or [latex]-\infty[/latex].

Proof

We provide a proof of this theorem in the special case when [latex]f, \, g, \, f^{\prime}[/latex], and [latex]g^{\prime}[/latex] are all continuous over an open interval containing [latex]a[/latex]. In that case, since [latex]\underset{x\to a}{\lim}f(x)=0=\underset{x\to a}{\lim}g(x)[/latex] and [latex]f[/latex] and [latex]g[/latex] are continuous at [latex]a[/latex], it follows that [latex]f(a)=0=g(a)[/latex]. Therefore,

[latex]\begin{array}{lllll} \underset{x\to a}{\lim}\frac{f(x)}{g(x)} & =\underset{x\to a}{\lim}\frac{f(x)-f(a)}{g(x)-g(a)} & & & \text{since} \, f(a)=0=g(a) \\ & =\underset{x\to a}{\lim}\frac{\frac{f(x)-f(a)}{x-a}}{\frac{g(x)-g(a)}{x-a}} & & & \text{multiplying numerator and denominator by} \, \frac{1}{x-a} \\ & =\frac{\underset{x\to a}{\lim}\frac{f(x)-f(a)}{x-a}}{\underset{x\to a}{\lim}\frac{g(x)-g(a)}{x-a}} & & & \text{limit of a quotient} \\ & =\frac{f^{\prime}(a)}{g^{\prime}(a)} & & & \text{definition of the derivative} \\ & =\frac{\underset{x\to a}{\lim}f^{\prime}(x)}{\underset{x\to a}{\lim}g^{\prime}(x)} & & & \text{continuity of} \, f^{\prime} \, \text{and} \, g^{\prime} \\ & =\underset{x\to a}{\lim}\frac{f^{\prime}(x)}{g^{\prime}(x)} & & & \text{limit of a quotient} \end{array}[/latex]

 

Note that L’Hôpital’s rule states we can calculate the limit of a quotient [latex]\frac{f}{g}[/latex] by considering the limit of the quotient of the derivatives [latex]\frac{f^{\prime}}{g^{\prime}}[/latex]. It is important to realize that we are not calculating the derivative of the quotient [latex]\frac{f}{g}[/latex].

[latex]_\blacksquare[/latex]

Example: Applying L’Hôpital’s Rule (0/0 Case)

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

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

Watch the following video to see the worked solution to Example: Applying L’Hôpital’s Rule (0/0 Case).

Try It

Evaluate [latex]\underset{x\to 0}{\lim}\dfrac{x}{\tan x}[/latex].

We can also use L’Hôpital’s rule to evaluate limits of quotients [latex]\frac{f(x)}{g(x)}[/latex] in which [latex]f(x)\to \pm \infty[/latex] and [latex]g(x)\to \pm \infty[/latex]. Limits of this form are classified as indeterminate forms of type [latex]\infty / \infty[/latex]. Again, note that we are not actually dividing [latex]\infty[/latex] by [latex]\infty[/latex]. Since [latex]\infty[/latex] is not a real number, that is impossible; rather, [latex]\infty / \infty[/latex] is used to represent a quotient of limits, each of which is [latex]\infty[/latex] or [latex]−\infty[/latex].

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.

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

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

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

Try It

Evaluate [latex]\underset{x\to \infty }{\lim}\dfrac{\ln x}{5x}[/latex]

As mentioned, L’Hôpital’s rule is an extremely useful tool for evaluating limits. It is important to remember, however, that to apply L’Hôpital’s rule to a quotient [latex]\frac{f(x)}{g(x)}[/latex], it is essential that the limit of [latex]\frac{f(x)}{g(x)}[/latex] be of the form [latex]0/0[/latex] or [latex]\infty / \infty[/latex] Consider the following example.

Example: When L’Hôpital’s Rule Does Not Apply

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.

 

Try It

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.

Watch the following video to see the worked solution to the above Try It.

Try It