L’Hôpital’s Rule for indeterminate powers

Click For Summary
The discussion centers on the application of L'Hôpital's Rule to evaluate limits involving indeterminate forms, specifically when approaching zero. Participants clarify the reasoning behind swapping the limit of a function with the limit of its logarithm, emphasizing the continuity of the logarithmic and exponential functions. It is concluded that if the limit of the logarithm is 4, the limit of the original expression is indeed e^4. The continuity of the logarithm is noted as a motivation for taking logs, but not essential for the proof. Overall, the exchange highlights the importance of understanding function continuity in limit evaluations.
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!
 
Physics news on Phys.org
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​

##\ ##
 
Last edited:
  • Like
Likes FactChecker and member 731016
Continuity of the function ln() at 1. Of course, the answer will have to be converted back, so 4 is not the final answer.
 
  • Like
Likes member 731016
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)).
 
  • Like
Likes member 731016, SammyS and FactChecker
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.
 
  • Like
Likes member 731016
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.
 
  • Like
Likes member 731016 and FactChecker
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.
 
  • Like
Likes member 731016
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.
 
  • Like
Likes member 731016