$\lim_{x \to c} {f(x)} = 0 \Rightarrow \lim_{x \to c} {1 \over f(x)} = \infty$ If $f$ is defined as a function of real variables to real values, and $c \in cl(Domain)$ as its limit value (i.e.  $\lim_{x \to c} {f(x)} = 0 $) how to prove that this implies: $\lim_{x \to c} {1 \over f(x)} = \infty$.
It seems logical that the values will be always bigger, but when tried to construct a contradiction using the y-creterion I stuck at: $\exists \epsilon > 0: f(x)>0 \forall x \in [c-\epsilon,c+\epsilon]$.
 A: This is problematic, even if you consider $1/|f(x)|$ instead of $1/f(x)$. For example, let $$f(x)=\begin{cases}x\sin(1/x) & x\ne0\\ 0 & \text{otherwise.}\end{cases}$$ This is everywhere defined and continuous on $\Bbb R$, and $$\lim_{x\to 0}f(x)=0,$$ but since there is no $x$-interval around $0$ on which $1/|f(x)|$ is defined, then it is problematic to talk about $$\lim_{x\to0}\frac1{|f(x)|}.$$ It's even more problematic to talk about it if we were to let $f$ be the constant zero function.
We must make some extra assumptions to take care of your problems. In particular, you need to show the following:
Suppose that $E\subseteq\Bbb R$ and $f:E\to\Bbb R.$ Let $F=\{x\in E:f(x)\ne0\}.$ Suppose further that $c\in\Bbb R$ is a limit point of both $E$ and $F,$ and that for all $\epsilon>0$ there is some $\delta>0$ such that $|f(x)|<\epsilon$ whenever $x\in E$ with $0<|x-c|<\delta$. Then for all $M,$ there exists $\delta>0$ such that $1/|f(x)|>M$ whenever $x\in F$ with $0<|x-c|<\delta.$
A: Let me just assume that $c = 0$. And as pointed out in the comment above, you have to be careful. Take the following just as an outline of a general approach. As pointed out in the excellent answer by Cameron, this doesn't always work. You have to make some assumptions. But maybe it can be helpful?
But, I assume that what want to prove that given $N$ there is an $\delta>0$ such that 
$$
\lvert x\rvert < \delta \Rightarrow \lvert f(x)^{-1}\rvert > N.
$$
So Let $N$ be given. Let $\epsilon = \frac{1}{N}$. Then there is a $\delta >0$ such that $\lvert f(x)\rvert < \epsilon = N^{-1}$.
That is you get exactly what you want because
$$\begin{align}
\lvert f(x)\rvert &< \epsilon = N^{-1} \quad \Rightarrow \\
\lvert f(x)\rvert^{-1} &> N.
\end{align}
$$
