# L’Hôpital’s Rule for indeterminate powers

• ChiralSuperfields
In summary: However, the limit of the logarithm at 1 is not interesting, as it is the same as the limit of the original expression. 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.

#### ChiralSuperfields

Homework Statement
Relevant Equations
For this,

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!

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
 x 1+sin(4x) ( )^cot x e^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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Last edited:
FactChecker and ChiralSuperfields
Continuity of the function ln() at 1. Of course, the answer will have to be converted back, so 4 is not the final answer.

ChiralSuperfields
I think it's continuity of exp we need, since we're not pulling the limit inside the log. $$\lim_{x \to a} f(x) = \lim_{x \to a} \exp(\ln f(x)) = \exp(\lim_{x \to a} \ln f(x)).$$

ChiralSuperfields, SammyS and FactChecker
pasmith said:
I think it's continuity of exp we need, since we're not pulling the limit inside the log. $$\lim_{x \to a} f(x) = \lim_{x \to a} \exp(\ln f(x)) = \exp(\lim_{x \to a} \ln f(x)).$$
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.

ChiralSuperfields
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 $a$.

ChiralSuperfields 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.

ChiralSuperfields
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.

ChiralSuperfields