prove if f(x) has an infinite limit then limit of 1/f(x) is = 0 I wanted to ask if someone can do me the favor pointing out the mistakes I might of made in proving the theorem below. Also is there a way to prove the theorem without using the definition of limits?
Theorem:
If $\lim_{x\to c}$f(x) =$\infty$ then $\lim_{x\to c} \frac{1}{f(x)}$ = $0$
Proof:
Rewriting the above statement using the definition of limits and definition of infinite limits:
($ 0<|x-c|<\delta \Rightarrow f(x)>M)$ $\Longrightarrow (0<|x-c|<\delta \Rightarrow |\frac{1}{f(x)} -0|< \epsilon $)
where $\epsilon$ and $M$ are both greater than zero and M denotes any real number.
Taking the second conditional statement:      ($0<|x-c|<\delta \Rightarrow |\frac{1}{f(x)} -0|< \epsilon $)
we can see that
(i) $$-\epsilon <\frac{1}{f(x)} <\epsilon$$   
Taking the first conditional statement:
 $ 0<|x-c|<\delta \Rightarrow f(x)>M$, we can conclude using the reciprocal of inequalities that $f(x)>M>0$ is equivalent to 
(ii)$$0<\frac{1}{f(x)}<\frac{1}{M}$$
From (i) and (ii) we get$$0<\frac{1}{f(x)}<\frac{1}{M}<\epsilon.$$
Meaning for any chosen value of $\epsilon>0$ and $M>0$ we can rewrite the theorem as 
($ 0<|x-c|<\delta \Rightarrow \frac{1}{f(x)}<\frac{1}{M}<\epsilon )$ $\Longrightarrow (0<|x-c|<\delta \Rightarrow |\frac{1}{f(x)} -0|< \epsilon $)
which shows that for any $\epsilon$ there exists a$\frac{1}{f(x)}$  that is always lower than $\frac{1}{M}$.
 A: I think you are lost in symbolism. It is better to use some amount of language. The starting statement which you used about $\delta, M, \epsilon$ is simply not making any sense.
What we need to show is the following statement:
If "(P) corresponding to any $M > 0$ we can find a $\delta > 0$ such that $f(x) > M$ whenever $0 < |x - c| < \delta$" then "(Q) corresponding to any $\epsilon > 0$ we can find a $\delta' > 0$ such that $|1/f(x)| < \epsilon$ whenever $0 < |x - c| < \delta'$".
This is a logical statement of type $P \Rightarrow Q$ where $P$ are $Q$ are marked clearly in italics in the last paragraph. These statements $P, Q$ can further be written using logical symbols. For example $P$ can be written as $\forall M > 0,\, \exists \delta > 0,\, 0 < |x - c| < \delta \Rightarrow f(x) > M$. Similarly we can write $Q$. But you haven't written in this manner which makes the statement very vague and confusing.
Another fundamental mistake is that you have used the same $\delta$ in both $P$ and $Q$. This is so very wrong. I have explicitly used $\delta'$ to make it clear that both $\delta$ and $\delta'$ have no relation with each other.
Why $P$ implies $Q$? This is obvious. For truth of $Q$ we need to find a $\delta'$ for an $\epsilon$. Let $M = 1/\epsilon$ and from the truth of $P$ we find a $\delta$ based on $M$ and set $\delta' = \delta$. This chosen $\delta'$ will ensure that $0 < |x - c| < \delta' \Rightarrow |1/f(x)| < \epsilon$.
I have used the above logical symbols only to align with your post. A better answer goes like this:
Let $\epsilon > 0$ be given and set $M = 1/\epsilon > 0$. Since $f(x) \to \infty$ as $x \to c$, it is possible to find a $\delta > 0$ such that $f(x) > M$ whenever $0 < |x - c| < \delta$. Thus $0 < 1/f(x) < \epsilon$ whenever $0 < |x - c| < \delta$. This implies that it is possible to find a $\delta > 0$ such that $|1/f(x)| < \epsilon$ whenever $0 < |x - c| < \delta$. Hence $1/f(x) \to 0$ as $x \to c$.
A: You are saying nowhere what's the relationship between $\varepsilon$ and $M$.
The proof is more straightforward.
Suppose $\varepsilon>0$. Then, by assumption, there exists $\delta>0$ such that, for all $x$ with $0<|x-c|<\delta$, we have $f(x)>1/\varepsilon$. This implies that
$$
\text{for all $x$, if $0<|x-c|<\delta$, then } 0<\frac{1}{f(x)}<\varepsilon
$$
Since $\varepsilon$ is arbitrary, we have proved that
$$
\lim_{x\to c}\frac{1}{f(x)}=0
$$
