Answer
See explanation
Work Step by Step
Because $\lim_{x\rightarrow c}=\infty$,
for every $M>0$, there is a $\delta>0$ so that if $0<|x-c|<\delta$, then $f(x)>M$.
Let's consider $\epsilon>0$. For $M=\frac{1}{\epsilon}$, there is $\delta>0$ so that if $0<|x-c|<\delta$, then $f(x)>M=\frac{1}{\epsilon}$.
But $f(x)>\frac{1}{\epsilon}$ is equivalent to $\frac{1}{f(x)}<\epsilon$.
Therefore $\lim_{x\rightarrow c}\frac{1}{f(x)}=0$
