Given: $\lim\limits_{n\to\infty}a_n=0$ prove $\lim\limits_{n\to\infty}\frac{1}{a_n}=\infty$ Please help me to prove $\displaystyle\lim_{n\to\infty}a_n=0$ then $\displaystyle\lim_{n\to\infty}\frac{1}{a_n}=\infty$
Please give me a hint, not a full solution.
I know how to prove $a_n\to\infty \Rightarrow \frac{1}{a_n}\to0$, but not the other way around.
The original problem:
Given $\forall a\in\left\{ a_n \right\}, a<0$ and $\displaystyle\lim_{n\to\infty}a_n=0$ prove: $\displaystyle\lim_{n\to\infty}\frac{1}{a_n}=-\infty$
 A: Since $\{a_n\}$ is negative and $a_n\to 0$, for each $M\in\mathbb{N}$ there exists $N$ such that $-\frac{1}{M}<a_n<0$ for all $n\geq N$.
Therefore $\frac{1}{a_n}<-M$ for all $n\geq N$, which implies that $\frac{1}{a_n}\to-\infty$.
A: Just which $\text{“}\infty\text{''}$ are we talking about?  In some contexts it makes sense to distinguish between $+\infty$ and $-\infty$, and in some other contexts it makes sense to talk about just one $\text{“}\infty\text{''}$ that's at both ends of the real line and is approached by going in either the positive or the negative direction.  For example, $\lim\limits_{x\to\pi/2} \tan x = \infty$, where this is the last-mentioned $\infty$.  It is this last-mentioned $\infty$ that is approached by the reciprocal of something that approaches $0$.
If $a_n$ is close to $0$, then $-\varepsilon < a_n < \varepsilon$.  That implies $a_n>1/\varepsilon$ or $a_n<-1/\varepsilon$.
To show $1/a_n$ approaches $\infty$, you need to show that no matter how big a number $N$ is, $1/a_n$ ultimately gets bigger than $N$ in absolute value, i.e. either bigger than $N$ or less than $-N$.
So let $\varepsilon = 1/N$.
However, if you want $1/a_n$ to approach $+\infty$ or $-\infty$, then you need $a_n$ to be always positive or always negative, respectively, with only finitely many exceptions.
