How to prove this formula. Formula is for evaluate limits which answers are e to the power something.

Question

I would like to ask you about how to prove this formula. I come up with this formula when i was browsing internet. It works when i used it on given examples, but i would like to know if it works in general and how to prove it. The formula is $$\lim_{x \to \infty} f(x)^{g(x)} = \lim_{x \to \infty} e^{g(x)\cdot(f(x)-1)}$$ Consider f is polynomial function of first degree. Thanks for answers.

Answer 1

This is not generally true:

Take $f \equiv2$ and $g \equiv3$ the left side is equal $8$, the right side is equal $e^3$.

Answer 2

This formula is true Only when $$\lim_{x \to \infty} f(x) = 1 \space \& \lim_{x \to \infty} g(x) \to \infty$$

Proof:

let $$L = \lim_{x \to \infty} f(x)^{g(x)}$$ Taking log on both sides, we get $$log(L) = \lim_{x \to \infty} g(x) \space log(f(x))$$ now, since $\lim_{a \to 0} \frac{log(1+a)}{a}=1$, if we substitute $1+a$ as $f(x)$, the limit changes to $\lim_{x \to \infty} \frac{log(f(x))}{f(x)-1}=1$ or $\lim_{x \to \infty} log(f(x)) = f(x)-1$

substituting the above result in the expression for $log(L)$, we obtain $$ log(L) = \lim_{x \to \infty} g(x) \cdot (f(x)-1)$$

On taking an exponent both sides and removing the log, we get $$L = \lim_{x \to \infty} e^{\space g(x) \cdot (f(x)-1)}$$ which is the required result