Using definition of limits 
Let $c∈\mathbb{R}$ and let $f:\mathbb{R}\setminus{c}\rightarrow\mathbb{R}$ be a function such that $f(x)>0$ for all $x∈\mathbb{R}$. Use the definition of limits to prove that
$$
\lim_{x\to c}f(x)=\infty  \space\space\space\space\space\space\space\space\space\space\space   \text{iff}  \space\space\space\space\space\space\space\space\space\space\space    \lim_{x\to c}\frac{1}{f(x)}=0.
$$

Proving the "$\Rightarrow$": Here is the definition: $\lim_{x\to c}f(x)=\infty$ if $\forall M∈\mathbb{R},\exists\delta>0$ such that $\forall x∈\mathbb{R}, 0<|x-c|<\delta\Rightarrow f(x)>M$. Here is my proof:
Let $\epsilon >0$ and set $M=\frac{1}{\epsilon}$. Since $\lim_{x\to c}f(x)=\infty$, we can find a $\epsilon >0$ such that $f(x)>M$ whenever $0<|x-c|<\delta$.
Thus $0<\frac{1}{f(x)}<\frac{1}{\epsilon}$ whenever $0<|x-c|<\delta$. This implies that it is possible to find a $\delta>0$ such that $|\frac{1}{f(x)}|<\epsilon$ whenever $0<|x-c|<\delta$. Since $\epsilon$ is arbitrary, we have proved that $\lim_{x\to c}\frac{1}{f(x)}=0$.
Proving the "$\Leftarrow$": This proof I am unsure of. I know that by the definition of a limit, $\lim_{x\to c} f(x) = 0$ if $\forall\epsilon>0, \exists\delta>0$ such that $\forall x∈\mathbb{R}\setminus{c}, 0<|x-c|<\delta \Rightarrow |f(x)-0|<\epsilon$. I am unsure of how to define $\lim_{x\to c}\frac{1}{f(x)}=0$ in a similar way. Any advice would be greatly appreciated.
 A: For the $\leftarrow$ direction. Let $M > 0$ be given, $\exists \delta > 0$ such that: $0 < |x-c| < \delta\implies \dfrac{1}{f(x)} < \dfrac{1}{M}\implies f(x) > M$. this shows that the limit is $\infty$ as claimed.
A: For you $\Longrightarrow$ direction, I have some critiques.

Let $\epsilon >0$ and set $M=\frac{1}{\epsilon}$. Since $\lim_{x\to c}f(x)=\infty$, we can find a $\epsilon >0$ such that $f(x)>M$ whenever $0<|x-c|<\delta$.

I think that this is a typo, but I believe that the third $\epsilon$ should be a $\delta$. Otherwise, this is wrong and doesn't seem to make sense.

Thus $0<\frac{1}{f(x)}<\frac{1}{\epsilon}$ whenever $0<|x-c|<\delta$.

This is incorrect. If you have $$f(x)>M=\frac{1}{\epsilon}$$ then you should end up with $$\epsilon>\frac{1}{f(x)}>0 $$whenever $0<|x-c|<\delta$
After that then your proof is fine.
For the $\Longleftarrow$ direction, I have some suggestions.
You say you have some confusion with writing out the definition of the limit for $1/f(x)$. Hopefully the following helps:
We have a function $f:\mathbb{R}\setminus\{c\}\to\mathbb{R}$. Define the function $g:\mathbb{R}\setminus\{c\}\to\mathbb{R}$ as $g(x)=1/f(x)$. This is well-defined since $f(x)>0$ for all $x\in\mathbb{R}\setminus\{c\}$. Now $$\lim_{x\to c}g(x)=\lim_{x\to c}\frac{1}{f(x)}$$ so, writing out the definition in terms of $g$, we have $$\forall\epsilon>0\,\exists\delta>0\,\forall x\in\mathbb{R}\setminus\{c\}\left(0<|x-c|<\delta\Longrightarrow |g(x)-0|<\epsilon\right)$$
Replace $g(x)$ with $1/f(x)$ to see how that definition works.
A: Both directions use the same concept.
$\implies$
$\lim\limits_{x\to c} f(x) = \infty$ so for any $\epsilon> 0$ and $M =\frac 1\epsilon > 0$ there is a $\delta$ so that $|x-c|< \delta \implies f(x) > M =\frac 1{\epsilon}$ so $|x-c| < \delta \implies 0 < \frac 1{f(x)} = |\frac 1{f(x)} -0| < \frac 1M =\epsilon$.
So $\lim\limits_{x\to c} \frac 1{f(x)} = 0$.
$\Leftarrow$
$\lim\limits_{x\to c}\frac 1{f(x)} =0$ means for every $M > 0$ and $\epsilon = \frac 1M > 0$ there is a $\delta > 0$ so that $ |x-c|< \delta\implies |\frac 1{f(x)} - 0|=\frac 1{f(x)} < \epsilon = \frac 1M$.
And as $f(x) > 0$ then $|x-c| < \delta \implies f(x) > \frac 1{\epsilon} = M$.
So $\lim\limits_{x\to c}\frac 1{f(x)} =\infty$.
.......
For the desired $\epsilon$ or $M$ you just use $\frac 1\epsilon, \frac 1M$ to show there is a $\delta$ so that $|x -c| \delta$ implies the limit you know for the reciprical of the value you want, and then by take the reciprical you get the limit of the reciprical fits the value you want.
The exact same procedure either way.
