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.
Best Answer
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.