There is a theorem which says "If $\lim\limits_{x\to a} f(x) =\lim\limits_{x\to a} g(x)=0$ such that $\lim\limits_{x\to a} \frac{f(x)}{g(x)}$ exists, then $\lim\limits_{x \to a} \{1+f(x)\}^{\frac{1}{g(x)}} = e^{\lim\limits_{x\to a} {\frac{f(x)}{g(x)}}}$
The proof for this theorem goes as such:
$$\text{Let } A=\lim_{x\to a} \{1+f(x)\}^{1\over g(x)}.\text{ Then, }$$
$$\log_e A = \lim_{x\to a} \frac{\log\{1+f(x)\}}{g(x)}$$
$$\implies \log_e A = \lim_{x\to a} \frac{\log\{1+f(x)\}}{f(x)} \cdot \frac{f(x)}{g(x)}$$
$$\implies \log_e A = \lim_{x\to a} \frac{f(x)}{g(x)}$$
$$\implies A=e^{\lim\limits_{x\to a} {\frac{f(x)}{g(x)}}}$$
Now my doubt here is… how did the first step become second step by taking log on both sides? How did the log go inside the limit? That is my confusion.
That is, $A=\lim_{x\to a} \{1+f(x)\}^{1\over g(x)}$ became $\log_e A = \lim_{x\to a} \frac{\log\{1+f(x)\}}{g(x)}$. How?
Hope you get what my doubt is. Thanks.
Best Answer
Regarding your question about the first step: $\ln$ is continuous, i.e. you can write the limit inside this function (given that it exists). The $g(x)$ in the denominator comes from the fact that $$\ln(a^b)=b\ln(a),$$ where you can choose $a=1+f(x)$ and $b=1/g(x)$.