Reciprocal of a limit that goes to infinity Lets say we have a limit $\lim_{n \rightarrow \infty} \frac{a_n}{b_n} = +\infty$, then is it safe to assume that $\lim_{n \rightarrow \infty} \frac{b_n}{a_n} = 0$?
 A: In general, yes, as $$\lim_{n \to \infty} \frac{b_n}{a_n} = \lim_{n \to\infty} \dfrac{1}{a_n\big/b_n}.$$ And in this particular case, since $a_n\big/b_n$ goes to $\infty$ as $n\to\infty$, it follows that the reciprocal will go to $0^+\!\!$.
A: We have that $\lim\limits_{n\to \infty}\dfrac{a_n}{b_n}$. By definition, this means that: 
$$\forall M>0 \, \exists N \mid n\geq N\implies  \frac{a_n}{b_n}>M\begin{aligned}\implies& \operatorname{sgn}(a_n)=\operatorname{sgn}(b_n)\implies 0<\frac{b_n}{a_n}<\frac1M\\ \implies& a_n\neq 0\end{aligned}.$$
Now for $\lim\limits_{n\to \infty}\dfrac{b_n}{a_n}$. If we are given some $\varepsilon>0$, then we can choose an $M$ such that $0<\dfrac1M<\varepsilon$. Once we have chosen this $M$, we know that there is an $N$, such that $n\geq N \implies \left\vert \dfrac{b_n}{a_n}\right \vert <\dfrac 1M<\varepsilon$ and $\dfrac{b_n}{a_n}>0$.
To sum up: we may say that $$\forall \varepsilon>0 \, \exists N \mid n\geq N\begin{aligned}\implies&  \left\vert\frac{b_n}{a_n}-0\right \vert<\varepsilon\\ \implies&\dfrac{b_n}{a_n}>0\end{aligned}.$$ This brings us to the conlusion that $\lim\limits_{n\to \infty}\dfrac{b_n}{a_n}=0$, or to  be more precise $$\frac{b_n}{a_n}\downarrow 0\text{ as } n\to\infty.$$
