Proof of $$\lim_{x\rightarrow \infty} f(x)^{g(x)} = c^d$$

Theorem:

$$c,\ d\in {\bf R},\ \lim_{x\rightarrow \infty} f(x)=c>0,\ \lim_{x\rightarrow \infty} g(x) =d>0$$

then

$$\lim_{x\rightarrow \infty} f(x)^{g(x)} = c^d$$

Proof: Because

1. $y(x)=ln(x)$ is continuous at $x = c > 0$

2. $\lim\limits_{x\to \infty}f(x) = c$

according to the composition law, we have

$$\lim\limits_{x \to \infty}lnf(x) = ln\lim\limits_{x \to \infty}f(x) = lnc$$

Because $\lim\limits_{x \to \infty}g(x) = d$, we have

$$\lim\limits_{x\to \infty}g(x)lnf(x) = \lim\limits_{x\to \infty}g(x)\cdot\lim\limits_{x \to \infty}lnf(x) = dlnc$$

Apply composition law again, we get

$$\lim\limits_{x\to \infty}f(x)^{g(x)} = \lim\limits_{x\to \infty}e^{g(x)lnf(x)} = e^{\lim\limits_{x\to \infty}g(x)lnf(x)} = e^{dlnc} = c^d$$