# A limit topic proof

1. Dec 31, 2014

### 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?

2. Dec 31, 2014

### Staff: 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.

3. Dec 31, 2014

### Raghav Gupta

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

4. Dec 31, 2014

### Staff: 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.

5. Dec 31, 2014

### Joshua L

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

6. Jan 1, 2015

### Raghav Gupta

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

7. Jan 1, 2015

### FactChecker

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.

8. Jan 2, 2015

### Raghav Gupta

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: May 7, 2017
9. Jan 2, 2015

### lurflurf

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

10. Jan 2, 2015

### Raghav Gupta

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

11. Jan 3, 2015

### Staff: Mentor

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

12. Jan 3, 2015

### Stephen Tashi

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

13. Jan 3, 2015

### Staff: Mentor

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.