Chapter 3: Applications of Differentiation
Section 3.9: Indeterminate Forms and L'Hôpital's Rule
Evaluate limx→0x+1−1x, then detail an applicable strategy taken from Table 3.9.1.
Evaluation of the Limit
Context Panel: Evaluate and Display Inline
limx→0x+1−1x = 12
Since x+1−1x tends to the indeterminate form 0/0 as x→0, apply L'Hôpital's rule to obtain
The typical reasoning at this point would be that as x→0, the denominator tends to 2⋅1=2. Hence, the limit is 1/2. Maple's stepwise solution takes greater pains, as seen below.
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
Context Panel: Student Calculus1≻All Solution Steps
limx→0x+1−1x→show solution stepsLimit Stepslimx→0⁡x+1−1x▫1. 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+1−1x=12⁢x+1This gives:limx→0⁡12⁢x+1▫2. Apply the constant multiple rule to the term limx→0⁡12⁢x+1◦Recall the definition of the constant multiple rulelimx→0⁡⁢f⁡x=⁢limx→0⁡f⁡x◦This means:limx→0⁡12⁢x+1=We can rewrite the limit as:limx→0⁡1x+12▫3. Apply the power rule◦Recall the definition of the power rulelimx→a⁡x=limx→a⁡xThis gives:12⁢limx→0⁡x+1▫4. Apply the sum rule◦Recall the definition of the sum rulelimx→a⁡f⁡x+g⁡x=limx→a⁡f⁡x+limx→a⁡g⁡xf⁡x=xg⁡x=1This gives:12⁢limx→0⁡x+limx→0⁡1▫5. Apply the constant rule to the term limx→0⁡1◦Recall the definition of the constant ruleLimit⁡C,x=C◦This meanslimx→0⁡1=1We can now rewrite the limit as:12⁢limx→0⁡x+1▫6. Apply the identity rule◦Recall the definition of the identity rulelimx→a⁡x=aThis gives:12
Notice how Maple first deals with the factor of 2, then carefully moves the limit inside the square root and applies the sum rule to x+1. All of these precise details are incorporated into the "reasoning" at the end of the Mathematical Solution in the previous subsection.
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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