Let $f$ and $g$ be real valued functions defined on $(a,b)$. If $\lim_{x \to a^+} f(x) $ exists in $\Bbb R$ and $\lim_{x \to a^+} g(x)= \infty $, prove that $\lim_{x \to a^+} \frac{f(x)}{g(x)} = 0$.
I have no idea how to approach this question can you guys give be some hint? is this kind of obvious to say $$\frac{\text{finite value}} \infty = 0?$$
What do you have to show it here?
Hint: Can you begin by proving the following fact about sequences?
If $a_n$ is a bounded sequence, and $b_n \to \infty$, then $$\lim_{n \to \infty} \frac{a_n}{b_n} = 0$$
Proceeding directly from the definitions is advised here.
One way to proceed with proving this is to suppose that $|a_n| < M$ for some positive $M$ and all $n$. Then choosing $n$ large enough guarantees that $|b_n| > \frac{M}{\epsilon}$ for any preassigned error $\epsilon$. Now consider $|a_n / b_n|$. This proof can be immediately translated into language about functions, using the fact that if the limit as $x \to a^+$ exists, then $f$ is bounded near $a$.
Suppose (without lose of generally) that $\lim_{x\to a^+} f(x)=b>0$. Then $$0\leq \lim_{x\to a^+} \frac{f(x)}{g(x)}\leq \lim_{x\to a^+} \frac{2b}{g(x)}=0.$$ And the result follows inmediately from the squeeze theorem.
