# A limit topic proof

This is a basic formula but can't find any proof of it. If anyone can explain or give a link showing proof it would be helpful.
## \lim_{x\rightarrow a} f(x)^{g(x)} ## = ##e^{\lim_{x\rightarrow a} g(x)[f(x)-1]}##
What is the proof for it?

mfb
Mentor
In general, this formula is wrong.
f(x)=2, g(x)=2. The left side has a limit of 4, the right side has a limit of e2 independent of a.

In general, this formula is wrong.
f(x)=2, g(x)=2. The left side has a limit of 4, the right side has a limit of e2 independent of a.
Yeah , I have made a mistake. It's true for f(a)=1 and g(a)= infinity. What is the proof then?

mfb
Mentor
g(a)= infinity does not make sense, assuming your functions are defined as ##\mathbb R \to \mathbb R##.
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.

Is this the formula that you are referring to? $$\lim_{x \rightarrow a} \ [ \ {f(x)^{g(x)}} \ ] = e^{\lim_{x \rightarrow a} {[ \ln {( \ f(x) \ )} g(x)} ] }$$

Is this the formula that you are referring to? $$\lim_{x \rightarrow a} \ [ \ {f(x)^{g(x)}} \ ] = e^{\lim_{x \rightarrow a} {[ \ln {( \ f(x) \ )} g(x)} ] }$$
No. This Is fairly simple the antilog would cancel the log to get the question back.
g(a)= infinity does not make sense, assuming your functions are defined as ##\mathbb R \to \mathbb R##.
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.
See the second method in the link. From where the formula came?
http://www.vitutor.com/calculus/limits/one_infinity.html

FactChecker
Gold Member
See the second method in the link. From where the formula came?
http://www.vitutor.com/calculus/limits/one_infinity.html
I hate to say it, but this link looks to me like it is full of nonsense. It pretends to be proving that 1 = e. I reject that out of hand. Maybe I am missing something.

I hate to say it, but this link looks to me like it is full of nonsense. It pretends to be proving that 1 = e. I reject that out of hand. Maybe I am missing something.
It is not absolute 1, that is equal to 1. But it is all about limits. It is actually 1.000…01, which maybe small value but not 1.
e has the definition like e=
##\left( 1 + \frac{c}{n}\right)^n## for ##n \to \infty## This also seems like 1 case.
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?

Last edited by a moderator:
lurflurf
Homework Helper
suppose
$$L=\lim_{x\rightarrow a}\mathrm{f}(x)^{\mathrm{g}(x)}\\ \lim_{x\rightarrow a}\mathrm{f}(x)=1\\ \lim_{x\rightarrow a}\mathrm{g}(x)=\infty\\ \text{clearly}\\ \mathrm{f}(x)^{\mathrm{g}(x)}=\exp\left(\mathrm{g}(x)[\mathrm{f}(x)-1]\dfrac{\log(\mathrm{f}(x))}{\mathrm{f}(x)-1}\right)\\ %\phantom{\mathrm{f}(x)^{\mathrm{g}(x)}} \text{since exp is continuous the limit may be moved inside}\\ L=\lim_{x\rightarrow a}\mathrm{f}(x)^{\mathrm{g}(x)}\\ \phantom{L}=\lim_{x\rightarrow a}\exp\left(\mathrm{g}(x)[\mathrm{f}(x)-1]\dfrac{\log(\mathrm{f}(x))}{\mathrm{f}(x)-1}\right)\\ \phantom{L}=\exp\left(\lim_{x\rightarrow a}\, \left\{ \mathrm{g}(x)[\mathrm{f}(x)-1]\dfrac{\log(\mathrm{f}(x))}{\mathrm{f}(x)-1}\right\} \right)\\ \phantom{L}=\exp\left( \left\{ \lim_{x\rightarrow a}\, \mathrm{g}(x)[\mathrm{f}(x)-1] \right\} \left\{ \lim_{x\rightarrow a}\,\dfrac{\log(\mathrm{f}(x))}{\mathrm{f}(x)-1}\right\} \right)\\ L=\exp\left( \lim_{x\rightarrow a}\, \mathrm{g}(x)[\mathrm{f}(x)-1] \right)\\ \text{we have used the product rule for limits and the obvious fact}\\ \lim_{x\rightarrow a}\,\dfrac{\log(\mathrm{f}(x))}{\mathrm{f}(x)-1}=1$$

suppose
$$L=\lim_{x\rightarrow a}\mathrm{f}(x)^{\mathrm{g}(x)}\\ \lim_{x\rightarrow a}\mathrm{f}(x)=1\\ \lim_{x\rightarrow a}\mathrm{g}(x)=\infty\\ \text{clearly}\\ \mathrm{f}(x)^{\mathrm{g}(x)}=\exp\left(\mathrm{g}(x)[\mathrm{f}(x)-1]\dfrac{\log(\mathrm{f}(x))}{\mathrm{f}(x)-1}\right)\\ %\phantom{\mathrm{f}(x)^{\mathrm{g}(x)}} \text{since exp is continuous the limit may be moved inside}\\ L=\lim_{x\rightarrow a}\mathrm{f}(x)^{\mathrm{g}(x)}\\ \phantom{L}=\lim_{x\rightarrow a}\exp\left(\mathrm{g}(x)[\mathrm{f}(x)-1]\dfrac{\log(\mathrm{f}(x))}{\mathrm{f}(x)-1}\right)\\ \phantom{L}=\exp\left(\lim_{x\rightarrow a}\, \left\{ \mathrm{g}(x)[\mathrm{f}(x)-1]\dfrac{\log(\mathrm{f}(x))}{\mathrm{f}(x)-1}\right\} \right)\\ \phantom{L}=\exp\left( \left\{ \lim_{x\rightarrow a}\, \mathrm{g}(x)[\mathrm{f}(x)-1] \right\} \left\{ \lim_{x\rightarrow a}\,\dfrac{\log(\mathrm{f}(x))}{\mathrm{f}(x)-1}\right\} \right)\\ L=\exp\left( \lim_{x\rightarrow a}\, \mathrm{g}(x)[\mathrm{f}(x)-1] \right)\\ \text{we have used the product rule for limits and the obvious fact}\\ \lim_{x\rightarrow a}\,\dfrac{\log(\mathrm{f}(x))}{\mathrm{f}(x)-1}=1$$
Got it lurflurf for explaining in wonderful way.Special thanks to you.Thanks to others also for replying.

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

Stephen Tashi