Say you have the following function:
$$\frac{a^x}{b^x}$$
and want to evaluate it's limit as $x$ approaches infinity. Intuition tells me that the limit will approach $0$ if $b > a$ and will approach $\infty$ if $a>b$ and plugging various equations into a graphing calculator bears this out.
However I am at a loss for how to definitively prove that this is the case. Is there a specific identity or theorem for evaluating such a limit?
Hint:
write it as: $$ \lim_{x \to \infty}\left(\frac{a}{b} \right)^x=\lim_{x \to \infty} e^{x (\ln a-\ln b)} $$
In THIS ANSWER, I showed using only the limit definition of the exponential function and Bernoulli's Inequality that the exponential function satisfies the inequalities
$$\bbox[5px,border:2px solid #C0A000]{1+x\le e^x \le \frac {1}{1-x}} \tag 1$$
for $x<1$.
Noting that $\frac{a^x}{b^x}=e^{x\log(a/b)}$, application of $(1)$ reveals that
$$\begin{align} \bbox[5px,border:2px solid #C0A000]{1+x\log(a/b)\le \frac{a^x}{b^x}\le \frac{1}{1-x\log(a/b)} }\tag 2 \end{align}$$
for $x\log(a/b)<1$.
If $a=b$, then we see that $\lim_{x\to \infty}\frac{a^x}{b^x}=1$.
If $a>b$, then the left-hand side inequality yields $\lim_{x\to \infty}\frac{a^x}{b^x}=\infty$.
If $a<b$, then the right-hand side inequality yields $\lim_{x\to \infty}\frac{a^x}{b^x}=0$.
