$\underset{x \to 0}{\lim}(1 + g(x))^\frac{1}{f(x)} = e^{\underset{x \to 0}{\lim}\frac {g(x)}{f(x)}}$.

Question

If $\underset{x \to 0}{\lim} f(x)=0$ and $\underset{x \to 0}{\lim} g(x)=0$

then how the following equality happens?

My Try : This Will happen when ${\underset{x \to 0}{\lim}\frac {g(x)}{f(x)}}$ is finite.

Can anyone please correct me if I have gone wrong anywhere?

Thank You In Advance.

This is my older question, it might be helpful: https://math.stackexchange.com/questions/2454273/how-to-prove-this-limit-properly — Botond, Aug 12 '18 at 16:06
Possible duplicate https://math.stackexchange.com/questions/1460918/is-the-formula-lim-limits-x-to-a-1fxgx-e-lim-limits-x-to-afx/ — rtybase, Aug 12 '18 at 16:23

score 0 · Answer 1 · answered Aug 12 '18 at 16:04

0

Hint: $\log(1+g(x))\sim g(x)$ as $x\rightarrow 0$, because $g(x)\rightarrow 0$.

answered Aug 12 '18 at 16:04

A. Pongrácz

7,418

$\log(1+g(x))\sim 0$ as $x\rightarrow 0$, because $g(x)\rightarrow 0$. Am I not correct?@A Pongracz – cmi Aug 12 '18 at 16:10
That is true also, but not useful in this case. Try it with my hint, it will work! – A. Pongrácz Aug 12 '18 at 16:17
Your hint will work I think only when $\underset{x \to 0}{\lim}\frac {g(x)}{f(x)}$ exists.@A Pongrac – cmi Aug 12 '18 at 16:31
for this function $(1 + x) ^{\frac {1} {x^2}}$ will your hint work?@A Pongrac – cmi Aug 12 '18 at 16:33
it works in general – A. Pongrácz Aug 12 '18 at 17:14
take the logrithm of both sides first, and my hint shows the exact reason immediately – A. Pongrácz Aug 12 '18 at 17:44

score 0 · Answer 2 · answered Aug 12 '18 at 18:54

Let $\lim_{x \to a}p(x) = 1$ and $\lim_{x \to a} h(x) = \infty$

$\lim_{x \to a}p(x)^{h(x)} = \lim_{x \to a}(1+p(x) - 1)^{h(x)} = \lim_{x \to a}(1+\frac{1}{\frac{1}{p(x) - 1}})^{h(x)} = \lim_{x \to a}(1+\frac{1}{\frac{1}{p(x) - 1}})^{\frac{h(x)(p(x) -1)}{p(x) - 1}} = \lim_{x \to a}e^{h(x)(p(x) - 1)}$

In our case : $p(x) = g(x) + 1$ and $h(x) = \frac{1}{f(x)}$

$\underset{x \to 0}{\lim}(1 + g(x))^\frac{1}{f(x)} = e^{\underset{x \to 0}{\lim}\frac {g(x)}{f(x)}}$.

2 Answers2