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

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?

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

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.

Answer 1

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

Answer 2

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 $h(x) = \frac{1}{f(x)}$