L’Hôpital’s Rule for indeterminate powers

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Homework Help Overview

The discussion revolves around the application of L'Hôpital's Rule in the context of evaluating limits involving indeterminate forms, specifically focusing on the behavior of logarithmic functions and their continuity when approaching limits. The original poster seeks clarification on the justification for swapping limits involving a logarithmic transformation.

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants explore the implications of continuity for logarithmic and exponential functions in limit evaluation. Questions arise regarding the validity of swapping limits and the necessary conditions for applying logarithmic transformations in the context of limits.

Discussion Status

The discussion is ongoing, with participants providing insights into the continuity of functions involved and the reasoning behind taking logarithms. There is an exploration of different interpretations regarding the limit process, but no explicit consensus has been reached.

Contextual Notes

Some participants note the importance of understanding the behavior of functions near specific points, particularly in relation to continuity and the application of L'Hôpital's Rule. The original poster's question highlights a potential gap in the explanation of the limit swapping process.

member 731016
Homework Statement
Please see below
Relevant Equations
Please see below
For this,
1684824210724.png

Does someone please know why we are allowed to swap the limit as x approaches zero from the right of y with that of In y?

Thank you for any help!
 
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Hi,
Isnt it so that if the limit of the logarithm is 4, then the limit the exercise asks for is ##e^4## ?
I don't see any swapping
x1+sin(4x)( )^cot xe^4
0.1​
1.389418​
26.51954​
54.59815​
0.01​
1.039989​
50.44658​
0.001​
1.004​
54.16361​
0.0001​
1.0004​
54.55449​
0.00001​
1.00004​
54.59378​

##\ ##
 
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Continuity of the function ln() at 1. Of course, the answer will have to be converted back, so 4 is not the final answer.
 
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I think it's continuity of exp we need, since we're not pulling the limit inside the log. <br /> \lim_{x \to a} f(x) = \lim_{x \to a} \exp(\ln f(x)) = \exp(\lim_{x \to a} \ln f(x)).
 
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pasmith said:
I think it's continuity of exp we need, since we're not pulling the limit inside the log. <br /> \lim_{x \to a} f(x) = \lim_{x \to a} \exp(\ln f(x)) = \exp(\lim_{x \to a} \ln f(x)).
Good point. But I think we need the continuity of the log at one point and the continuity of exp at the end to undo what was done with the log.
 
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Clearly it doesn't help us to look at g\left(\lim_{x \to a} g^{-1}(f(x))\right) if we can't say anything about the behaviour of g^{-1}\circ f near a.
 
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pasmith said:
Clearly it doesn't help us to look at g\left(\lim_{x \to a} g^{-1}(f(x))\right) if we can't say anything about the behaviour of g^{-1}\circ f near a.
I see your point. Although the continuity of ln() is good motivation for taking logs, it is not an essential part of the proof.
 
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BvU said:
Isnt it so that if the limit of the logarithm is 4, then the limit the exercise asks for is e4?
I don't see any swapping
The swapping occurs in going from ##\lim_{x \to 0^+} \ln y = \dots = 4## to the next equation which isn't shown. Namely, ##\lim_{x \to 0^+} \ln y = \ln(\lim_{x \to 0^+} <\text{stuff}>) = 4##. Here the limit operation and natural log are swapped.
From this we conclude that the original expression has a limit of ##e^4##.
FactChecker said:
Although the continuity of ln() is good motivation for taking logs, it is not an essential part of the proof.
Taking logs of both sides is the standard way of dealing with the limits of exponential expressions where the variable occurs in the exponent and possibly elsewhere.
 
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