Let $ I \subset \mathbb{R}$ be an open interval and let $c \in I$ and $f:I-\left \{ c \right \} \rightarrow \mathbb{R}$ be a function. Suppose that $f(x) \neq 0:\forall x \in I-\left \{ c \right \}$ , and that $\lim_{x \rightarrow c}f(x)=0$. Prove that $\lim_{x \rightarrow c} \frac{1}{f(x)}$ does not exist.
Proof
Suppose we assume that the limit exists. Then, $$ \lim_{x \rightarrow c}f(x)= mc+b $$ $$ \implies \lim_{x \rightarrow c}\frac{1}{f(x)}= \frac{1}{mc+b} $$ Assume we have $f(x)=3x+2$ as our test function. So, $$ \lim_{x \rightarrow c} \frac{1}{f(x)}=\frac{1}{3c+2} $$
Do I need to find the positive and left limit for both sides? I'm confused on how to proceed?
Let $C>0$ be given. Then, choose $\epsilon >0$ such that $$ \frac{1}{\epsilon} \geq C $$ Since $\lim_{x \to c}f(x)=0$, we can find $\delta >0$ such that $|x-c| <\delta$ implies $|f(x)| < \epsilon$. This implies: $$ \frac{1}{|f(x)|} > \frac{1}{\epsilon} \geq C $$ Thus, $\lim_{x \to c} \frac{1}{|f(x)|}= \pm \infty$ .
Can you take it from there?