# Proof for Limit Topic Formula: e^[g(x)[f(x)-1]]

• Raghav Gupta
In summary: The hypothesis that excludes the case when f(x) is the constant function f(x) = 1 is that the limit of f(x) as x approaches a is equal to 1. This is necessary for the proof to work because it allows for the use of the logarithmic function, which is only defined for positive values. If f(x) is a constant function that is equal to 1, then the logarithmic function cannot be used and the proof would not hold.

#### Raghav Gupta

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?

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.

mfb said:
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?

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? $$\lim_{x \rightarrow a} \ [ \ {f(x)^{g(x)}} \ ] = e^{\lim_{x \rightarrow a} {[ \ln {( \ f(x) \ )} g(x)} ] }$$

Joshua L said:
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.
mfb said:
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

Raghav Gupta said:
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.

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

lurflurf said:
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.

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

mfb said:
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 ?

Then the whole manipulation made above does not work, but this special case is easy to handle separately.
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.

## What is the "Proof for Limit Topic Formula: e^[g(x)[f(x)-1]]"?

The "Proof for Limit Topic Formula: e^[g(x)[f(x)-1]]" is a mathematical formula used to find the limit of a function as it approaches a specific value. It is commonly used in calculus and involves the use of the natural number e.

## How is the "Proof for Limit Topic Formula: e^[g(x)[f(x)-1]]" derived?

The "Proof for Limit Topic Formula: e^[g(x)[f(x)-1]]" is derived from the definition of the natural logarithm and the properties of limits. It involves manipulating the original function using algebraic and logarithmic rules to simplify the expression and find the limit.

## What are the main properties of the "Proof for Limit Topic Formula: e^[g(x)[f(x)-1]]"?

The main properties of the "Proof for Limit Topic Formula: e^[g(x)[f(x)-1]]" are that it can be used to find the limit of a function as it approaches a specific value, and it involves the natural number e, which is approximately equal to 2.71828. It also requires the use of logarithms and algebraic manipulation to simplify the expression.

## What are some common applications of the "Proof for Limit Topic Formula: e^[g(x)[f(x)-1]]"?

The "Proof for Limit Topic Formula: e^[g(x)[f(x)-1]]" is commonly used in calculus to find the limit of a function as it approaches a specific value. It is also used in other areas of mathematics and science, such as physics and engineering, to model and analyze various phenomena.

## What are some tips for using the "Proof for Limit Topic Formula: e^[g(x)[f(x)-1]]" effectively?

To use the "Proof for Limit Topic Formula: e^[g(x)[f(x)-1]]" effectively, it is important to have a strong understanding of logarithmic and algebraic rules and how to apply them. It is also helpful to practice with various examples to gain a better understanding of the formula and its applications. Additionally, double-checking calculations and using a graphing calculator or software can also aid in effectively using this formula.

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