4.5 Indeterminate Forms and L’Hôpital’s Rule

298

OBJECTIVES

When you finish this section, you should be able to:

  1. Identify indeterminate forms of the type \(\dfrac{0}{0}\) and \(\dfrac{\infty }{\infty }\) (p. 298)
  2. Use L’Hôpital’s Rule to find a limit (p. 299)
  3. Find the limit of an indeterminate form of the type \(0\cdot \infty \), \(\infty -\infty \), \(0^{0}\), \(1^{\infty }\), or \(\infty ^{0}\) (p. 303)

In this section, we reexamine the limit of a quotient and explore what we can do when previous limit theorems cannot be used.

NEED TO REVIEW?

Properties of limits are discussed in Section 1.2, pp. 87-88.

1 Identify Indeterminate Forms of the Type 0/0 and /∞

To find \(\lim\limits_{x\rightarrow c}\dfrac{f(x) }{g(x) }\) we usually first try to use the Limit of a Quotient: \[ \begin{equation*} \lim\limits_{x\rightarrow c}\dfrac{f(x) }{g(x) }= \dfrac{\lim\limits_{x\rightarrow c}f(x) }{\lim\limits_{x\rightarrow c}g(x) } \tag{1} \end{equation*} \]

But this result cannot always be used. For example, to find \( \lim\limits_{x\rightarrow 2}\dfrac{x^{2}-4}{x-2},\) we cannot use equation (1), since the numerator and the denominator each approach \(0\) (resulting in the form \(\dfrac{0}{0}).\) Instead, we use algebra and obtain \[ \lim\limits_{x\rightarrow 2}\dfrac{x^{2}-4}{x-2}=\lim\limits_{x\rightarrow 2} \dfrac{( x-2) ( x+2) }{x-2}=\lim\limits_{x\rightarrow 2}( x+2) =4 \]

NEED TO REVIEW?

Limits at infinity and infinite limits are discussed in Section 1.5, pp. 117-125.

To find \(\lim\limits_{x\rightarrow \infty }\dfrac{3x-2}{x+5}\), we cannot use equation (1), since the numerator and the denominator each become unbounded (resulting in the form \(\dfrac{\infty }{\infty }\)). Instead, we divide the numerator and denominator by \(x.\) Then \[ \lim\limits_{x\rightarrow \infty }\dfrac{3x-2}{x+5}=\lim\limits_{x\rightarrow \infty }\dfrac{3-\dfrac{2}{x}}{1+\dfrac{5}{x}}=\dfrac{\lim\limits_{x\rightarrow \infty }\left( 3-\dfrac{2}{x}\right) }{\lim\limits_{x\rightarrow \infty }\left( 1+\dfrac{5}{x}\right) }=3 \]

As a third example, to find \(\lim\limits_{x\rightarrow 0}\dfrac{\sin x}{x}\), we cannot use equation (1) since it leads to the form \(\dfrac{0}{0}\). Instead, we use a geometric argument (the Squeeze Theorem) to show that \[ \lim\limits_{x\rightarrow 0}\dfrac{\sin x}{x}=1 \]

NOTE

The word “indeterminate” conveys the idea that the limit cannot be found without additional work.

Whenever using \(\lim\limits_{x\rightarrow c}\dfrac{f(x) }{g(x) }=\dfrac{\lim\limits_{x\rightarrow c}f(x) }{\lim\limits_{x\rightarrow c}g(x) }\) leads to the form \(\dfrac{0}{0}\) or \(\dfrac{\infty }{\infty }\), we say that \(\dfrac{f}{g}\) is an indeterminate form at \(c\). There are other indeterminate forms that we discuss in Objective 3.

299

spanDEFINITIONspan Indeterminate Form at \(c\) of the Type \(\dfrac{0}{0}\) or the Type \(\dfrac{\infty }{\infty }\)

If the functions \(f\) and \(g\) are each defined in an open interval containing the number \(c,\) except possibly at \(c\), then the quotient \(\dfrac{f(x) }{g(x) }\) is called an indeterminate form at \(c\) of the type \(\dfrac{0}{ 0}\) if \[ \lim\limits_{x\rightarrow c}f(x) =0\qquad \hbox{and} \qquad \lim\limits_{x\rightarrow c}g(x) =0 \]

and an indeterminate form at \( c\) of the type \(\dfrac{\infty }{\infty }\) if \[ \lim\limits_{x\rightarrow c}f(x) =\pm \infty\qquad \hbox{and}\qquad \lim\limits_{x\rightarrow c}g(x) =\pm \infty \]

NOTE

\(\dfrac{0}{0}\) and \(\dfrac{\infty }{\infty }\) are symbols used to denote an indeterminate form.

These definitions also hold for limits at infinity.

Identifying Indeterminate Forms of the Types \(\dfrac{0}{0}\) and \(\dfrac{\infty }{\infty }\)

  1. \(\dfrac{\cos ( 3x) -1}{2x}\) is an indeterminate form at \(0\) of the type \(\dfrac{0}{0}\) since \[\lim\limits_{x\rightarrow 0}[ \cos ( 3x) -1] =0\, \hbox{and}\, \lim\limits_{x\rightarrow 0}\,(2x) =0\]
  2. \(\dfrac{x-1}{x^{2}+2x-3}\) is an indeterminate form at \(1\) of the type \(\dfrac{0}{0}\) since \[ \hbox{\(\lim\limits_{x\rightarrow 1}( x-1) =0\) and \(\lim\limits_{x\rightarrow 1}( x^{2}+2x-3) =0\)} \]
  3. \(\dfrac{x^{2}-2}{x-3}\) is not an indeterminate form at \(3\) of the type \(\dfrac{0}{0}\) since \(\lim\limits_{x\rightarrow 3}( x^{2}-2) \neq 0\).
  4. \(\dfrac{x^{2}}{e^{x}}\) is an indeterminate form at \(\infty \) of the type \(\dfrac{\infty }{\infty }\) since \[ \hbox{\(\lim\limits_{x\rightarrow \infty}x^{2}=\infty \) and \(\lim\limits_{x\rightarrow \infty }e^{x}=\infty \)} \]

NOW WORK

Problem 7.

2 Use L’Hôpital’s Rule to Find a Limit

A theorem, named after the French mathematician Guillaume François de L’Hôpital (pronounced “low-p-tal”), provides a method for finding the limit of an indeterminate form.

THEOREM L’Hôpital’s Rule

Suppose the functions \(f\) and \(g\) are differentiable on an open interval \(I\) containing the number \(c\), except possibly at \(c,\) and \(g^\prime (x) \neq 0\) for all \(x\neq c\) in \(I\). Let \(L\) denote either a real number or \(\pm \infty \), and suppose \(\dfrac{f(x) }{g(x) }\) is an indeterminate form at \(c\) of the type \(\dfrac{0}{0}\) or \( \dfrac{\infty }{\infty }\). If \(\lim\limits_{x\rightarrow c}\dfrac{f^\prime (x) }{g^\prime (x) }=L\), then \(\lim\limits_{x\rightarrow c}\dfrac{f(x) }{g(x) }=L.\)

L’Hôpital’s Rule is also valid for limits at infinity and one-sided limits. A partial proof of L’Hôpital’s rule is given in Appendix B. A limited proof is given here that assumes \(f^{\prime}\) and \(g^{\prime}\) are continuous at \(c\) and \(g^\prime(c)\ne 0\).

300

Proof

Suppose \(\lim\limits_{x\rightarrow c}f(x) =0\) and \(\lim\limits_{x\rightarrow c}g(x) =0.\) Then \(f( c) =0\) and \(g( c) =0.\) Since both \(f^\prime \) and \(g^\prime \) are continuous at \(c\) and \(g^\prime ( c) \neq 0,\) we have \[ \begin{eqnarray*} &&\lim\limits_{x\rightarrow c}\dfrac{f^\prime (x) }{g^\prime (x) } = \dfrac{\lim\limits_{x\rightarrow c}f^\prime (x) }{ \lim\limits_{x\rightarrow c}g^\prime (x) }= \dfrac{ f^\prime ( c) }{g^\prime ( c) }=\dfrac{ \lim\limits_{x\rightarrow c}\dfrac{f(x) -f( c) }{x-c}}{\lim\limits_{x\rightarrow c}\dfrac{g(x) -g( c) }{x-c}}\\[-21.8pt] &&\hspace{5.6pc}\underset{\color{#0066A7}{{\hbox{Quotient}}\atop{\hbox{Property}}}} {\color{#0066A7}{{\uparrow}}} \quad\hspace{28pt} \underset{ \color{#0066A7}{{\hbox{$f^\prime ,\,\,g^\prime$} }\atop{\hbox{continuous}}}} {\color{#0066A7}{\uparrow}} \quad\kern-6pt \underset{\color{#0066A7}{\hbox{Definition of}}\atop\color{#0066A7}{\hbox {a derivative}}}{\hspace{-1pc}{\hspace{-12pt}{\color{#0066A7}{\uparrow}}\hspace{12pt}}}\\[5pt] &&\hspace{7pc}=\lim\limits_{x\rightarrow c}\dfrac{\dfrac{f(x) -f( c) }{x-c}}{\dfrac{g(x) -g( c) }{x-c}}=\lim\limits_{x\rightarrow c}\dfrac{f(x) }{g(x) } \\[-21.7pt] &&\hspace{19pc}\underset{\color{#0066A7}{\hbox{$f( c) =0$}}\atop\color{#0066A7}{\hbox{$g( c) =0$}}}{\color{#0066A7}{\uparrow}} \end{eqnarray*} \]

ORIGINS

Guillaume François de L’Hôpital (1661–1704) was a French nobleman. When he was 30, he hired Johann Bernoulli to tutor him in calculus. Several years later, he entered into a deal with Bernoulli. L’Hôpital paid Bernoulli an annual sum for Bernoulli to share his mathematical discoveries with him but no one else. In 1696 L’Hôpital published the first textbook on differential calculus. It was immensely popular, the last edition being published in 1781 (seventy-seven years after L’Hôpital’s death). The book included the rule we study here. After L’Hôpital’s death, Johann Bernoulli made his deal with L’Hôpital public and claimed that L’Hôpital’s textbook was his own material. His position was dismissed because he often made such claims. However, in 1921, the manuscripts of Bernoulli’s lectures to L’Hôpital were found, showing L’Hôpital’s calculus book was indeed largely Johann Bernoulli’s work.

Steps for Finding a Limit Using L’Hôpital’s Rule

Step 1 Check that \(\dfrac{f}{g}\) is an indeterminate form at \(c\) of the type \(\dfrac{0}{0}\) or \(\dfrac{\infty }{\infty }\). If it is not, do not use L’Hôpital’s Rule.

Step 2 Differentiate \(f\) and \(g\) separately.

Step 3 Find \(\lim\limits_{x\rightarrow c}\dfrac{f^\prime (x) }{g^\prime (x) }\). This limit is equal to \(\lim\limits_{x\rightarrow c}\dfrac{f(x) }{g(x) }\), provided the limit is a number or \(\infty \) or \(-\infty\).

Step 4 If \(\dfrac{f^\prime }{g^\prime }\) is an indeterminate form at \(c\), repeat the process.

Using L’Hôpital’s Rule to Find a Limit

Find \(\lim\limits_{x\rightarrow 0}\dfrac{\tan x}{6x}\).

Solution We follow the steps for finding a limit using L’Hôpital’s Rule.

Step 1 Since \(\lim\limits_{x\rightarrow 0}\tan x=0\) and \(\lim\limits_{x\rightarrow 0}( 6x) =0\), the quotient \(\dfrac{\tan x }{6x}\) is an indeterminate form at \(0\) of the type \(\dfrac{0}{0}\).

Step 2 \(\dfrac{d}{\textit{dx}}\tan x=\sec ^{2}x\) and \(\dfrac{d}{\textit{dx}}\left( 6x\right) =6\).

Step 3 \(\lim\limits_{x\rightarrow 0}\dfrac{\dfrac{d}{\textit{dx}}\tan x}{\dfrac{d}{dx}\left( 6x\right) }=\) \(\lim\limits_{x\rightarrow 0}\dfrac{\sec ^{2}x}{6}=\dfrac{1}{6}.\)

It follows from L’Hôpital’s Rule that \(\lim\limits_{x\rightarrow 0}\dfrac{\tan x}{6x}=\dfrac{1}{6}\).

CAUTION

When using L’Hôpital’s Rule, we find the derivative of the numerator and the derivative of the denominator separately. Be sure not to find the derivative of the quotient. Also, using L’Hôpital’s Rule to find a limit of an expression that is not an indeterminate form may result in an incorrect answer.

In the solution of Example 2, we were careful to determine that the limit of the ratio of the derivatives, that is, \(\lim\limits_{x\rightarrow 0}\dfrac{\sec ^{2}x}{6}\), existed or became infinite before using L’Hôpital’s Rule. However, the usual practice is to combine Step 2 and Step 3 as follows: \[ \begin{eqnarray*} \lim\limits_{x\rightarrow 0}\dfrac{\tan x}{6x} &=& \lim\limits_{x\rightarrow 0} \dfrac{\dfrac{d}{\textit{dx}}\tan\;x}{\dfrac{d}{\textit{dx}}( 6x) }= \lim\limits_{x\rightarrow 0}\dfrac{\sec ^{2}\;x}{6}=\dfrac{1}{6} \end{eqnarray*} \]

At times, it is necessary to use L’Hôpital’s Rule more than once.

301

Using L’Hôpital’s Rule to Find a Limit

Find \(\lim\limits_{x\rightarrow 0}\dfrac{\sin x-x}{x^{2}}\).

Solution We use the steps for finding a limit using L’Hôpital’s Rule.

Step 1 Since \(\lim\limits_{x\rightarrow 0}\left( \sin x-x\right) =0\) and \(\lim\limits_{x\rightarrow 0}x^{2}=0\), the expression \(\dfrac{\sin x-x}{x^{2}}\) is an indeterminate form at \(0\) of the type \(\dfrac{0}{0}.\)

Steps 2 and 3 We use L’Hôpital’s Rule. \[ \begin{eqnarray*} &&\lim\limits_{x\rightarrow 0}\dfrac{\sin x-x}{x^{2}}=\lim\limits_{x\rightarrow 0}\dfrac{ \dfrac{d}{\textit{dx}}\left( \sin x-x\right) }{\dfrac{d}{\textit{dx}}x^{2}}=\lim\limits_{x \rightarrow 0}\dfrac{\cos x-1}{2x}=\dfrac{1}{2}\lim\limits_{x\rightarrow 0} \dfrac{\cos x-1}{x}\\ &&\hspace{8pc}\color{#0066A7}{\underset{\begin{array}{c}\hbox{L'H$\hat{\rm o}$pital's}\\ \hbox{Rule}\end{array}}{{\uparrow }}} \end{eqnarray*} \]

Since \(\lim\limits_{x\rightarrow 0}( \cos x-1) =0\) and \(\lim\limits_{x\rightarrow 0}x=0\), the expression \(\dfrac{\cos x-1}{x}\) is an indeterminate form at \(0\) of the type \(\dfrac{0}{0}\). So, we use L’Hôpital’s Rule again. \[ \begin{eqnarray*} &&\lim\limits_{x\rightarrow 0}\dfrac{\sin x-x}{x^{2}}=\dfrac{1}{2} \lim\limits_{x\rightarrow 0}\dfrac{\cos\;x-1}{x}=\dfrac{1}{2}\lim\limits_{x\rightarrow 0}\dfrac{\dfrac{d}{\textit{dx}}( \cos\;x-1) }{\dfrac{d}{\textit{dx}}x}=\dfrac{1}{2} \lim\limits_{x\rightarrow 0}\dfrac{-\!\sin\;x}{1}=0\\[-1.8pc] &&\hspace{17pc}\color{#0066A7}{\underset{\begin{array}{c}\hbox{L'H$\hat{\rm o}$pital's}\\ \hbox{Rule}\end{array}}{{\uparrow }} } \end{eqnarray*} \]

NOW WORK

Problem 29.

Using L’Hôpital’s Rule to Find a Limit at Infinity

Find:

  1. \(\lim\limits_{x\rightarrow \infty }\dfrac{\ln x}{x}\)
  2. \(\lim\limits_{x\rightarrow \infty }\dfrac{x}{e^{x}}\)
  3. \(\lim\limits_{x\rightarrow \infty }\dfrac{e^{x}}{x}\)

Solution (a) Since \(\lim\limits_{x\rightarrow \infty }\ln x=\infty \) and \(\lim\limits_{x\rightarrow \infty }x=\infty \), \(\dfrac{\ln x}{x}\) is an indeterminate form at \(\infty \) of the type \(\dfrac{\infty }{\infty }\). Using L’Hôpital’s Rule, we have \[ \begin{eqnarray*} &&\lim\limits_{x\rightarrow \infty }\dfrac{\ln x}{x} = \lim\limits_{x\rightarrow \infty } \dfrac{\dfrac{d}{\textit{dx}}\ln x}{\dfrac{d}{\textit{dx}}x}=\lim\limits_{x\rightarrow \infty } \dfrac{\dfrac{1}{x}}{1}=\lim\limits_{x\rightarrow \infty }\dfrac{1}{x}=0\\[-1.8pc] &&\hspace{3.35pc}\color{#0066A7}{\underset{\begin{array}{c}\hbox{L'H$\hat{\rm o}$pital's}\\ \hbox{Rule}\end{array}}{{\uparrow }}} \end{eqnarray*} \]

(b) \(\lim\limits_{x\rightarrow \infty }x=\infty \) and \(\lim\limits_{x\rightarrow \infty }e^{x}=\infty \), so \(\dfrac{x}{e^{x}}\) is an indeterminate form at \(\infty \) of the type \(\dfrac{\infty }{\infty }\). Using L’Hôpital’s Rule, we have \[ \begin{eqnarray*} &&\lim\limits_{x\rightarrow \infty }\dfrac{x}{e^{x}}=\lim\limits_{x\rightarrow \infty } \dfrac{\dfrac{d}{\textit{dx}}x}{\dfrac{d}{\textit{dx}}e^{x}}=\lim\limits_{x\rightarrow \infty } \dfrac{1}{e^{x}}=0\\[-1.8pc] &&\hspace{2.5pc}\color{#0066A7}{\underset{\begin{array}{c}\hbox{L'H$\hat{\rm o}$pital's}\\ \hbox{Rule}\end{array}}{{\uparrow }}} \end{eqnarray*} \]

(c) From (b), we know that \(\dfrac{e^{x}}{x}\) is an indeterminate form at \(\infty \) of the type \(\dfrac{\infty }{\infty }\). Using L’Hô pital’s Rule, we have \[ \begin{eqnarray*} &&\lim\limits_{x\rightarrow \infty }\dfrac{e^{x}}{x}=\lim\limits_{x\rightarrow \infty } \dfrac{\dfrac{d}{\textit{dx}}e^{x}}{\dfrac{d}{\textit{dx}}x}=\lim\limits_{x\rightarrow \infty } \dfrac{e^{x}}{1}=\infty \\[-1.8pc] &&\hspace{2.5pc}\color{#0066A7}{\underset{ \begin{array}{c}\hbox{L'H$\hat{\rm o}$pital's}\\ \hbox{Rule}\end{array} }{{\uparrow }}} \end{eqnarray*} \]

NOW WORK

Problem 35.

302

The results from Example 4 tell us that \(y=x\) grows faster than \(y=\ln x\), and that \(y=e^{x}\) grows faster than \(y=x\). In fact, these inequalities are true for all positive powers of \(x\). That is, \[ {\bbox[#FAF8ED,5pt]{\bbox[5px, border:1px solid black, #F9F7ED]{ \lim\limits_{x\rightarrow \infty } \dfrac{\ln{x}}{x^{n}}=0\qquad \lim\limits_{x\rightarrow \infty }\dfrac{x^{n}}{e^{x}}=0 }}} \]

for \(n \ge 1\) an integer. You are asked to verify these results in Problems 95 and 96.

Sometimes simplifying first reduces the effort needed to find the limit.

Using L’Hôpital’s Rule to Find a Limit

Find \(\lim\limits_{x\rightarrow 0}\dfrac{\tan x-\!\sin x}{x^{2}\tan x}\).

Solution \(\dfrac{\tan x-\!\sin x}{x^{2}\tan x}\) is an indeterminate form at \(0\) of the type \(\dfrac{0}{0}\). We simplify the expression before using L’Hôpital’s Rule. Then it is easier to find the limit. \[ \begin{eqnarray*} \dfrac{\tan x-\!\sin x}{x^{2}\tan x}&=&\dfrac{\dfrac{\sin x}{\cos x}-\!\sin x}{x^{2}\cdot \dfrac{\sin x}{\cos x}}=\dfrac{\dfrac{\sin x-\!\sin x\cos x}{\cos x}}{\dfrac{x^{2}\sin x}{\cos x}}=\dfrac{\sin x\left( 1-\cos x\right) }{x^{2}\sin x}=\dfrac{1-\cos x}{x^{2}} \end{eqnarray*} \]

Since \(\dfrac{1-\cos x}{x^{2}}\) is an indeterminate form at \(0\) of the type \(\dfrac{0}{0},\) we use L’Hôpital’s Rule.

\[ \begin{eqnarray*} &&\lim\limits_{x\rightarrow 0}\dfrac{\tan x-\!\sin x}{x^{2}\tan x} =\lim\limits_{x\rightarrow 0}\dfrac{1-\cos x}{x^{2}}=\lim\limits_{x\rightarrow 0}\dfrac{ \dfrac{d}{\textit{dx}}\left( 1-\cos x\right) }{\dfrac{d}{\textit{dx}}x^{2}}=\lim\limits_{x \rightarrow 0}\dfrac{\sin x}{2x}\\[-20.7pt] &&\hspace{16.7pc}\color{#0066A7}{\underset{\hbox{L'H$\hat{\rm o}$pital's Rule}}{{\uparrow }}}\\[6pt] &&\hspace{9.7pc}=\dfrac{1}{2}\lim\limits_{x\rightarrow 0}\dfrac{\sin x}{x}=\dfrac{1}{2}\qquad \color{#0066A7}{{\hbox{$\lim\limits_{x\rightarrow 0}\dfrac{\sin x}{x} =1$}} } \end{eqnarray*} \]

Compare this solution to one that uses L’Hôpital’s Rule at the start.

CAUTION

Be sure to verify that a simplified form is an indeterminate form at \(c\) before using L’Hôpital’s Rule.

In Example 6, we use L’Hôpital’s Rule to find a one-sided limit.

Using L’Hôpital’s Rule to Find a One-Sided Limit

Find \(\lim\limits_{x\rightarrow 0^{+}}\dfrac{\cot x}{\ln x}\).

Solution Since \(\lim\limits_{x\rightarrow 0^{+}}\cot x=\infty \) and \(\lim\limits_{x\rightarrow 0^{+}}\ln x=-\infty \), \(\dfrac{\cot x}{\ln x}\) is an indeterminate form at \(0^{+}\) of the type \(\dfrac{\infty }{\infty }\). Using L’Hôpital’s Rule, we find \[ \begin{eqnarray*} &&\lim\limits_{x\rightarrow 0^{+}}\dfrac{\cot x}{\ln x}=\lim\limits_{x\rightarrow 0^{+}} \dfrac{\dfrac{d}{\textit{dx}}\cot x}{\dfrac{d}{\textit{dx}}\ln x}=\lim\limits_{x\rightarrow 0^{+}}\dfrac{-\csc ^{2}x}{\dfrac{1}{x}}=-\lim\limits_{x\rightarrow 0^{+}}\dfrac{x}{ \sin ^{2}x}=-\lim\limits_{x\rightarrow 0^{+}}\dfrac{\dfrac{d}{\textit{dx}}x}{\dfrac{d}{\textit{dx}}\sin ^{2}x}\\[-21pt] &&\hspace{4pc}\color{#0066A7}{\underset{\hbox{L'H$\hat{\rm o}$pital's Rule}} {{\kern1pt\left\uparrow{\vphantom{\vrule width0pc height11pt depth0pt}}\right.}}}\quad \ \hspace{10.5pc} \color{#0066A7}{\underset{\hbox{$\csc ^{2}x=\dfrac{1}{\sin^{2}x}$}}{{\kern1pt\left\uparrow{\vphantom{\vrule width0pc height11pt depth0pt}}\right.}}}\quad \, \hspace{4.5pt}\color{#0066A7}{\underset{\hbox{L'H$\hat{\rm o}$pital's Rule}}{{\kern1pt\left\uparrow{\vphantom{\vrule width0pc height11pt depth0pt}}\right.}}}\\ &&\hspace{7pc}=-\lim\limits_{x\rightarrow 0^{+}}\dfrac{1}{2\sin x\cos x}=\left( -\dfrac{1}{2}\right) \left( \lim\limits_{x\rightarrow 0^{+}}\dfrac{1}{\sin x}\right) \left( \lim\limits_{x\rightarrow 0^{+}}\dfrac{1}{\cos x}\right)\\[7pt] &&\hspace{7pc}=\left( -\dfrac{1}{2}\right) \left( \lim\limits_{x\rightarrow 0^{+}}\dfrac{1}{\sin x}\right) ( 1) =-\infty \end{eqnarray*} \]

NOW WORK

Problem 61.

3 Find the Limit of an Indeterminate Form of the Type 0 · , - , 00, 1, or ∞0

303

L’Hôpital’s Rule can only be used for indeterminate forms of the types \( \dfrac{0}{0}\) and \(\dfrac{\infty }{\infty }\). Indeterminate forms of the type \(0\cdot \infty \), \(\infty -\infty \), \(0^{0}\), \(1^{\infty }\), or \(\infty ^{0}\) need to be rewritten in one of the forms \(\dfrac{0}{0}\) or \(\dfrac{\infty }{\infty }\) before we can use L’Hôpital’s Rule to find the limit.

NOTE

It might be tempting to argue that \(0\cdot \infty\) is \(0\), since “anything” times \(0\) is \(0\), but \(0\cdot \infty\) is not a product of numbers. That is, \(0\cdot \infty \) is not “zero times infinity” it symbolizes “a quantity tending to zero” times “a quantity tending to infinity.”

Indeterminate Forms of the Type \(0\cdot \infty \)

Suppose \(\lim\limits_{x\rightarrow c}f(x) =0\) and \(\lim\limits_{x\rightarrow c}g(x) =\infty .\) Then the product \(f\cdot g\) is an indeterminate form at \(c\) of the type \(0\cdot \infty \). To find \(\lim\limits_{x\rightarrow c} (f\cdot g),\) we rewrite the product \(f \cdot g\) as one of the following quotients: \[ f\cdot g= \dfrac{f}{\dfrac{1}{g}}\qquad\hbox{or}\qquad f\cdot g= \dfrac{g}{\dfrac{1}{f}} \]

The right side of the equation on the left is an indeterminate form at \(c\) of the type \(\dfrac{0}{0}\) ; the right side of the equation on the right is of the type \(\dfrac{\infty }{\infty }\). Then we use L’Hôpital’s Rule with one of these equations, choosing the one for which the derivatives are easier to find. If our first choice does not work, then we try the other one.

Finding the Limit of an Indeterminate Form of the Type \(0\cdot \infty \)

Find:

  1. \(\lim\limits_{x\rightarrow 0^{+}}( x\ln x)\)
  2. \(\lim\limits_{x\rightarrow \infty }\left( x \ \sin \dfrac{1}{x}\right)\)

NOTE

We choose to use \(\dfrac{\ln x}{\dfrac{1}{x}}\) rather than \(\dfrac{x}{\dfrac{1}{\ln x}}\) because it is easier to find the derivatives of \(\ln x\) and \(\dfrac{1}{x}\) than it is to find the derivatives of \(x\) and \(\dfrac{1}{\ln x}.\)

Solution (a) Since \(\lim\limits_{x\rightarrow 0^{+}}x=0\) and \(\lim\limits_{x\rightarrow 0^{+}}\ln x=-\infty \), then \(x\ln x\) is an indeterminate form at \(0^{+}\) of the type \(0\cdot \infty \). We change \(x\ln x \) to an indeterminate form of the type \(\dfrac{\infty }{\infty }\) by writing \(x\ln x= \dfrac{\ln x}{\dfrac{1}{x}}\) and using L’Hôpital’s Rule. \[ \begin{eqnarray*} \lim\limits_{x\rightarrow 0^{+}}( x\ln x) &=&\lim\limits_{x\rightarrow 0^{+}}\dfrac{\ln x}{\dfrac{1}{x}}=\lim\limits_{x\rightarrow 0^{+}}\dfrac{\dfrac{d}{\textit{dx}}\ln x}{\dfrac{d}{\textit{dx}}\dfrac{1}{x}} =\lim\limits_{x\rightarrow 0^{+}}\dfrac{\dfrac{1}{x}}{-\dfrac{1}{x^{2}}}= \lim\limits_{x\rightarrow 0^{+}}( -x) =0\\[-20.9pt] &&\hspace{16pt}\quad\color{#0066A7}{\underset{\hbox{L'H$\hat{\rm o}$pital's Rule}} {{\kern1pt\left\uparrow{\vphantom{\vrule width0pc height13pt depth0pt}}\right.}}}\qquad\hspace{4.6pc} \hspace{3.7pc}\color{#0066A7}{\underset{\hbox{Simplify}}{{\kern1pt\left\uparrow{\vphantom{\vrule width0pc height13pt depth0pt}}\right.}}} \end{eqnarray*} \]

(b) Since \(\lim\limits_{x\rightarrow \infty }\;x=\infty \) and \(\lim\limits_{x\rightarrow \infty }\;\sin \dfrac{1}{x}=0\), then \(x\sin \dfrac{1}{x}\) is an indeterminate form at \(\infty \) of the type \(0\cdot \infty \). We change \(x\;\sin \dfrac{1}{x}\) to an indeterminate form of the type \(\dfrac{0}{0}\) byeak writing \[ \begin{eqnarray*} \lim\limits_{x\rightarrow \infty }x\;\sin \dfrac{1}{x}&=&\lim\limits_{x\rightarrow \infty } \dfrac{\sin \dfrac{1}{x}}{\dfrac{1}{x}}=\lim\limits_{t\rightarrow 0^{+}}\dfrac{\sin\;t}{t}=1 \\[-1.75pc] &&\qquad\hspace{26.1pt}\color{#0066A7}{\underset{{\hbox {Let} \;t=\dfrac{1}{x}}}{{\kern1pt\left\uparrow{\vphantom{\vrule width0pc height11pt depth0pt}}\right.}}} \end{eqnarray*} \]

NOTE

Notice in the solution to (b) that we did not need to use L’Hôpital’s Rule.

304

NOW WORK

Problem 45.

Indeterminate Forms of the Type \(\infty - \infty\)*

If the limit of a function results in the indeterminate form \(\infty -\infty \), it is generally possible to rewrite the function as an indeterminate form of the type \(\dfrac{0}{0}\) or \(\dfrac{\infty }{\infty }\) by using algebra or trigonometry.

Finding the Limit of an Indeterminate Form of the Type \(\infty -\infty \)

Find \(\lim\limits_{x\rightarrow 0^{+}}\left( \dfrac{1}{x}-\dfrac{1}{\sin x}\right)\).

Solution Since \(\lim\limits_{x\rightarrow 0^{+}}\dfrac{1}{x} =\infty\) and \(\lim\limits_{x\rightarrow 0^{+}}\dfrac{1}{\sin x}=\infty\), then \(\dfrac{1}{x}-\dfrac{1}{\sin\;x}\) is an indeterminate form at \(0^{+}\) of the type \(\infty -\infty.\) We rewrite the difference as a single fraction. \[ \lim\limits_{x\rightarrow 0^{+}}\left( \dfrac{1}{x}-\dfrac{1}{\sin\;x}\right) =\lim\limits_{x\rightarrow 0^{+}}\dfrac{\sin\;x-x}{x\sin\;x} \]

Then, \(\dfrac{\sin\;x-x}{x\sin\;x}\) is an indeterminate form at \(0^{+}\) of the type \(\dfrac{0}{0}\). Now we can use L’Hôpital’s Rule. \[ \begin{eqnarray*} &&\lim\limits_{x\rightarrow 0^{+}}\left( \dfrac{1}{x}-\dfrac{1}{\sin\;x}\right) =\lim\limits_{x\rightarrow 0^{+}}\dfrac{\sin\;x-x}{x\sin\;x}=\lim\limits_{x\rightarrow 0^{+}} \dfrac{\dfrac{d}{\textit{dx}}\left(\sin\;x-x\right) }{\dfrac{d}{\textit{dx}}\left(x\;\sin x\right)}=\lim\limits_{x\rightarrow 0^{+}}\dfrac{\cos\;x-1}{x\cos\;x+\sin\;x}\\[-20.7pt] &&\qquad\hspace{14.5pc}\color{#0066A7}{\underset{{\hbox{L'H$\hat{\rm o}$pital's Rule}}}{{\kern1pt\left\uparrow{\vphantom{\vrule width0pc height11pt depth0pt}}\right.}}}\\ &&\hspace{10pc}=\lim\limits_{x\rightarrow 0^{+}}\dfrac{\dfrac{d}{\textit{dx}} \left(\cos\;x-1\right) }{\dfrac{d}{\textit{dx}}\left(x\;\cos\;x+\;\sin\;x\right) } =\lim\limits_{x\rightarrow 0^{+}}\dfrac{-\!\sin\;x}{\left(-x\;\sin\;x+\cos\; x\right) +\cos\;x}\\[-20.8pt] &&\hspace{3.5pc}\color{#0066A7}{\underset{\hbox{Type $\dfrac{0}{0}$; use L'H$\hat{\rm o}$pital's Rule}}{{\kern1pt\left\uparrow{\vphantom{\vrule width0pc height11pt depth0pt}}\right.}}}\\[-6pt] &&\hspace{10pc}=\lim\limits_{x\rightarrow 0^{+}}\dfrac{\sin\;x}{x\sin\;x-2\cos\;x}=\dfrac{0}{-2}=0 \end{eqnarray*} \]

NOW WORK

Problem 47.

*The indeterminate form of the type \(\infty -\infty\) is a convenient notation for any of the following: \(\infty -\infty\), \(-\infty - ( -\infty )\), \(\infty + ( -\infty ) \). Note that \(\infty +\infty =\infty\) and \(( -\infty ) + ( -\infty ) =-\infty \) are not indeterminate forms.

Indeterminate Forms of the Type \(1^{\infty }\), \(0^{0}\), or \(\infty ^{0}\)

NEED TO REVIEW?

Properties of logarithms are discussed in Appendix A.1, pp. A-10 to A-11.

A function of the form \([ f(x) ] ^{g(x) }\) may result in an indeterminate form of the type \(1^{\infty }\), \(0^{0}\), or \(\infty ^{0}\). To find the limit of such a function, we let \(y=[ f(x) ] ^{g(x) }\) and take the natural logarithm of each side. \[ \ln\;y=\ln [ f(x) ] ^{g(x)}=g(x)\;\ln\;f(x) \]

The expression on the right will then be an indeterminate form of the type \(0\cdot \infty \) and we find the limit using the method of Example 7.

Steps for Finding \({\lim\limits_{x\rightarrow c} [\ f(x) ] ^{g(x)}}\) when \({[\ f(x)] ^{g(x)}}\) is an Indeterminate Form at \(c\), of the Type \({1^{\infty}}\), \({0^{0}}\), or \({\infty ^{0}}\)

Step 1 Let \(y=[ f(x) ] ^{g(x)}\) and take the natural logarithm of each side, obtaining \(\ln y=g(x)\;\ln f(x) .\)

Step 2 Find \(\lim\limits_{x\rightarrow c}\;\ln y\).

Step 3 If \(\lim\limits_{x\rightarrow c}\;\ln y=L\), then \(\lim\limits_{x\rightarrow c}y=e^{L}\).

305

These steps also can be used for limits at infinity and for one-sided limits.

Finding the Limit of an Indeterminate Form of the Type \(0\vphantom{1}^0\)

Find \(\lim\limits_{x\rightarrow 0^{+}}x^{x}\)

Solution The expression \(x^{x}\) is an indeterminate form at \(0^{+}\) of the type \(0^{0}.\) We follow the steps for finding \(\lim\limits_{x\rightarrow c}[ f(x) ] ^{g(x) }\).

CAUTION

Do not stop after finding \(\lim\limits_{x\rightarrow c} \ln\;{y} { (=L)}\). Remember, we want to find \(\lim\limits_{x\rightarrow c}{ y}(=e^{L}) .\)

Step 1 Let \(y=x^{x}.\) Then \(\ \ln y=x\ln x.\)

Step 2 \(\lim\limits_{x\rightarrow 0^{+}}\;\ln y=\lim\limits_{x\rightarrow 0^{+}}( x\;\ln x) =0\) [from Example 7(a)].

Step 3 Since \(\lim\limits_{x\rightarrow 0^{+}}\;\ln y=0\), \(\lim\limits_{x\rightarrow 0^{+}}y=e^{0}=1\).

NOW WORK

Problem 51.

Finding the Limit of an Indeterminate Form of the Type \(1^{\infty}\)

Find \(\lim\limits_{x\rightarrow 0^{+}}( 1+x) ^{1/x}\).

Solution The expression \(( 1+x) ^{1/x}\) is an indeterminate form at \(0^{+}\) of the type \(1^{\infty }\).

Step 1 Let \(y=( 1+x) ^{1/x}.\) Then \(\ln y=\dfrac{1}{x}\;\ln (1+x).\)

Step 2 \[\begin{array}{l} &\lim\limits_{x\rightarrow 0^{+}}\;\ln y=\lim\limits_{x\rightarrow 0^{+}}\dfrac{\ln ( 1+x) }{x}=\lim\limits_{x\rightarrow 0^{+}}\dfrac{\dfrac{d}{\textit{dx}}\;\ln \left( 1+x\right) } {\dfrac{d}{\textit{dx}}x}=\lim\limits_{x\rightarrow 0^{+}}\dfrac{\dfrac{1}{1+x}}{1}=1\\ &\hspace{10pc}\hspace{-.7pt}\color{#0066A7}{\underset{\hbox{Type $\dfrac{0}{0}$; use L'H$\hat{\rm o}$pital's Rule}}{{\kern1pt\left\uparrow{\vphantom{\vrule width0pc height11pt depth0pt}}\right.}}}\\ \end{array}\]

Step 3 Since \(\lim\limits_{x\rightarrow 0^{+}}\ln y=1\), \(\lim\limits_{x\rightarrow 0^{+}}y=e^{1}=e\).

NOW WORK

Problem 85.