L’Hôpital’s Rule for indeterminate powers

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SUMMARY

The discussion centers on the application of L'Hôpital's Rule and the continuity of logarithmic and exponential functions in evaluating limits. Participants confirm that if the limit of the logarithm of a function approaches 4, then the limit of the original function is e4. The key point is the ability to swap the limit and logarithm due to the continuity of the natural logarithm and the exponential function, which allows for the transformation of the limit expression. This swapping is crucial for simplifying the evaluation of indeterminate forms.

PREREQUISITES
  • Understanding of L'Hôpital's Rule
  • Knowledge of limits and continuity in calculus
  • Familiarity with logarithmic and exponential functions
  • Basic skills in manipulating indeterminate forms
NEXT STEPS
  • Study the application of L'Hôpital's Rule in various indeterminate forms
  • Learn about the continuity properties of logarithmic and exponential functions
  • Explore advanced limit techniques, including epsilon-delta definitions
  • Review examples of limits involving exponential expressions and logarithms
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Students and educators in calculus, mathematicians dealing with limits, and anyone seeking to understand the application of L'Hôpital's Rule in evaluating indeterminate forms.

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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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