If $\lim f(x) = 0,$ then $\lim 1/|f(x)| = \infty.$ The Problem: 

Suppose that $f:D\to\Bbb R$, where $D$ is a subset of $\Bbb R$ and $a$ is an accumulation point of $D$, $\lim_{x\to a}f(x)=0$, and $f(x)\ne0$ for any $x$ in $D$ in some neighborhood of $a$.  Prove that $$\lim_{x\to a}\frac{1}{|f(x)|}= \infty$$

Right away, I noticed a similarity between a theorem involving sequences which stated that if $\lim_{n→∞}a_n=∞$, then $\lim_{n→∞}1/a_n=0$.  Another useful theorem: $\lim_{x→a}f(x)=L$ if and only if for every sequence $(x_n)_n$ in the domain $D$, $\lim_{n→∞}x_n=a$, where $a$ is a cluster point.  
I could use some help trying to string these ideas together. 
 A: You almost want to start with the consequent/conclusion. That is, we want to show that $1/|f(x)|$ converges to infinity. To show that, we unwrap the definition of this as, for any sequence $a_n\to a$, for any $M$, there is $N$ such that for all $n>N$, we have $1/|f(a_n)|>M$. 
Well, to prove this, we know that for any such sequence, we can guarantee that for any such sequence, and for any $\epsilon=1/M$, there is $N$ so that any $n>N$ gives $|f(a_n)|<\epsilon=1/M$. That shouldn't be hard to prove. 
Some problems with this: 


*

*I gather your theorem is for $L$ finite, not $L=\infty$. 

*Your definition of converging to infinity may not involve sequences. 

*Your theorem about the reciprocal of a limit is going the wrong way, that is, it says $P\implies Q$ when you need something more like $Q\implies P$. 


It is probably better to approach it with a straight forward $\epsilon-\delta$ proof. But you are right in that using the sequence version of checking limits is often easier to deal with. 
