- #1

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## \lim_{x\rightarrow a} f(x)^{g(x)} ## = ##e^{\lim_{x\rightarrow a} g(x)[f(x)-1]}##

What is the proof for it?

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- Thread starter Raghav Gupta
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- #1

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## \lim_{x\rightarrow a} f(x)^{g(x)} ## = ##e^{\lim_{x\rightarrow a} g(x)[f(x)-1]}##

What is the proof for it?

- #2

mfb

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f(x)=2, g(x)=2. The left side has a limit of 4, the right side has a limit of e

- #3

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Yeah , I have made a mistake. It's true for f(a)=1 and g(a)= infinity. What is the proof then?

f(x)=2, g(x)=2. The left side has a limit of 4, the right side has a limit of e^{2}independent of a.

- #4

mfb

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Also, the function value exactly at "a" is not relevant for the limit.

I guess you mean those expressions as the limiting values.

This is more general than the limit of ##\left( 1 + \frac{c}{n}\right)^n## for ##n \to \infty##, but I guess the same proof can be used with some modifications.

- #5

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- #6

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No. This Is fairly simple the antilog would cancel the log to get the question back.

See the second method in the link. From where the formula came?

Also, the function value exactly at "a" is not relevant for the limit.

I guess you mean those expressions as the limiting values.

This is more general than the limit of ##\left( 1 + \frac{c}{n}\right)^n## for ##n \to \infty##, but I guess the same proof can be used with some modifications.

http://www.vitutor.com/calculus/limits/one_infinity.html

- #7

FactChecker

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I hate to say it, but this link looks to me like it is full of nonsense. It pretends to be proving that 1See the second method in the link. From where the formula came?

http://www.vitutor.com/calculus/limits/one_infinity.html

- #8

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It is not absolute 1I hate to say it, but this link looks to me like it is full of nonsense. It pretends to be proving that 1^{∞}= e^{⅓}. I reject that out of hand. Maybe I am missing something.

e has the definition like e=

##\left( 1 + \frac{c}{n}\right)^n## for ##n \to \infty## This also seems like 1

See this link's example 4.This is a better link than previous although the formula is not given for direct answer

http://pages.uoregon.edu/jcomes/253limits.pdf [Broken]

When I apply my formula here it is applicable

So what's the proof?

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- #9

lurflurf

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$$L=\lim_{x\rightarrow a}\mathrm{f}(x)^{\mathrm{g}(x)}\\

\lim_{x\rightarrow a}\mathrm{f}(x)=1\\

\lim_{x\rightarrow a}\mathrm{g}(x)=\infty\\

\text{clearly}\\

\mathrm{f}(x)^{\mathrm{g}(x)}=\exp\left(\mathrm{g}(x)[\mathrm{f}(x)-1]\dfrac{\log(\mathrm{f}(x))}{\mathrm{f}(x)-1}\right)\\

%\phantom{\mathrm{f}(x)^{\mathrm{g}(x)}}

\text{since exp is continuous the limit may be moved inside}\\

L=\lim_{x\rightarrow a}\mathrm{f}(x)^{\mathrm{g}(x)}\\

\phantom{L}=\lim_{x\rightarrow a}\exp\left(\mathrm{g}(x)[\mathrm{f}(x)-1]\dfrac{\log(\mathrm{f}(x))}{\mathrm{f}(x)-1}\right)\\

\phantom{L}=\exp\left(\lim_{x\rightarrow a}\, \left\{ \mathrm{g}(x)[\mathrm{f}(x)-1]\dfrac{\log(\mathrm{f}(x))}{\mathrm{f}(x)-1}\right\} \right)\\

\phantom{L}=\exp\left( \left\{ \lim_{x\rightarrow a}\, \mathrm{g}(x)[\mathrm{f}(x)-1] \right\} \left\{ \lim_{x\rightarrow a}\,\dfrac{\log(\mathrm{f}(x))}{\mathrm{f}(x)-1}\right\} \right)\\

L=\exp\left( \lim_{x\rightarrow a}\, \mathrm{g}(x)[\mathrm{f}(x)-1] \right)\\

\text{we have used the product rule for limits and the obvious fact}\\

\lim_{x\rightarrow a}\,\dfrac{\log(\mathrm{f}(x))}{\mathrm{f}(x)-1}=1

$$

- #10

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Got it lurflurf for explaining in wonderful way.Special thanks to you.Thanks to others also for replying.

$$L=\lim_{x\rightarrow a}\mathrm{f}(x)^{\mathrm{g}(x)}\\

\lim_{x\rightarrow a}\mathrm{f}(x)=1\\

\lim_{x\rightarrow a}\mathrm{g}(x)=\infty\\

\text{clearly}\\

\mathrm{f}(x)^{\mathrm{g}(x)}=\exp\left(\mathrm{g}(x)[\mathrm{f}(x)-1]\dfrac{\log(\mathrm{f}(x))}{\mathrm{f}(x)-1}\right)\\

%\phantom{\mathrm{f}(x)^{\mathrm{g}(x)}}

\text{since exp is continuous the limit may be moved inside}\\

L=\lim_{x\rightarrow a}\mathrm{f}(x)^{\mathrm{g}(x)}\\

\phantom{L}=\lim_{x\rightarrow a}\exp\left(\mathrm{g}(x)[\mathrm{f}(x)-1]\dfrac{\log(\mathrm{f}(x))}{\mathrm{f}(x)-1}\right)\\

\phantom{L}=\exp\left(\lim_{x\rightarrow a}\, \left\{ \mathrm{g}(x)[\mathrm{f}(x)-1]\dfrac{\log(\mathrm{f}(x))}{\mathrm{f}(x)-1}\right\} \right)\\

\phantom{L}=\exp\left( \left\{ \lim_{x\rightarrow a}\, \mathrm{g}(x)[\mathrm{f}(x)-1] \right\} \left\{ \lim_{x\rightarrow a}\,\dfrac{\log(\mathrm{f}(x))}{\mathrm{f}(x)-1}\right\} \right)\\

L=\exp\left( \lim_{x\rightarrow a}\, \mathrm{g}(x)[\mathrm{f}(x)-1] \right)\\

\text{we have used the product rule for limits and the obvious fact}\\

\lim_{x\rightarrow a}\,\dfrac{\log(\mathrm{f}(x))}{\mathrm{f}(x)-1}=1

$$

- #11

mfb

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I don't think the last fact is so obvious, but it follows from l'Hospital.

- #12

Stephen Tashi

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I don't think the last fact is so obvious, but it follows from l'Hospital.

What hypothesis excludes the case when f(x) is the constant function f(x) = 1 ?

- #13

mfb

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To cover everything, you can split the reals into two sets, one where f(x)=1 (where the formula is trivial and exact even without limit) and one where f(x) != 1 (where we can use the steps made in post 9) and combine them afterwards.

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