Chapter 3: Applications of Differentiation
Section 3.9: Indeterminate Forms and L'Hôpital's Rule
Evaluate limx→0+−lnxx, then detail an applicable strategy taken from Table 3.9.1.
As x→0+, −lnxx tends to the indeterminate form ∞0.
From the graph of −lnxx in Figure 3.9.10(a), it would appear that the required limit is 1. Indeed,
limx→0+−lnxx = 1
Compute the limit of the logarithm:
Figure 3.9.10(a) Graph of −lnxx
Hence, limx→0+−lnxx=e0=1, where, having computed the limit of the log, the limit of the log must then be exponentiated. Alternatively,
where, of course, the limit of x ln−lnx in the exponential is computed via L'Hôpital's rule. This is the path Maple's annotated stepwise solutions follow, namely, first a rewrite of the expression as the exponential of its log, with a passage of the limit from outside the exponential to the inside. Then, all the work of computing the limit must be done up in the exponential.
Annotated Stepwise Maple Solution
The Context Panel provides access to the options All Solutions, Next Step, and Limit Rules, as per the figure to the right.
While the Context Panel lists both a "difference" and a "sum" rule, Maple treats a difference as if it were a sum, thereby making no essential distinction between the two rules. Thus, where Table 1.3.1 distinguishes between a Sum and a Difference rule, Maple considers both to be the single Sum rule.
In addition, the Student Calculus1 package contains a ShowSolution command that can be applied to the inert form of the limit operator. The limit operator can be converted to the inert form through the Context Panel by selecting the options 2-D Math≻Convert To≻Inert Form.
Annotated stepwise solution via the Context Panel
Tools≻Load Package: Student Calculus 1
Calculus palette: Limit template
Context Panel: Student Calculus1≻All Solution Steps
limx→0+−lnxx→show solution stepsLimit Stepslimx→0+⁡−ln⁡xx▫1. Rewrite◦Equivalent expression−ln⁡xx=ⅇx⁢ln⁡−ln⁡xThis gives:limx→0+⁡ⅇx⁢ln⁡−ln⁡x▫2. Apply the exponential rule◦Recall the definition of the exponential rulelimx→a⁡ef⁡x=elimx→a⁡f⁡xThis gives:ⅇlimx→0+⁡x⁢ln⁡−ln⁡x▫3. Apply the L'Hôpital's Rule rule◦Recall the definition of the L'Hôpital's Rule rulelimx→c⁡f⁡xg⁡x=limx→c⁡ⅆⅆxf⁡xⅆⅆxg⁡x◦Rule appliedx⁢ln⁡−ln⁡x=−xln⁡xThis gives:ⅇlimx→0+⁡−xln⁡x▫4. Apply the constant multiple rule to the term limx→0⁡−xln⁡x◦Recall the definition of the constant multiple rulelimx→0⁡⁢f⁡x=⁢limx→0⁡f⁡x◦This means:limx→0⁡−xln⁡x=We can rewrite the limit as:ⅇ−limx→0+⁡xln⁡x▫5. Apply the quotient rule◦Recall the definition of the quotient rulelimx→a⁡f⁡xg⁡x=limx→a⁡f⁡xlimx→a⁡g⁡xf⁡x=xg⁡x=ln⁡xThis gives:ⅇ−limx→0+⁡xlimx→0+⁡ln⁡x▫6. Apply the identity rule◦Recall the definition of the identity rulelimx→a⁡x=aThis gives:1
This solution, or a shorter one, can be obtained with the
tutor. Selecting, for example, the Constant Multiple rule in the menu "Understood Rules" will suppress the step in which that rule is explicitly invoked.
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