lugreget9

2022-01-14

The proof that if f(x) is continuous at $x=c$, then $\frac{1}{f\left(x\right)}$ is continuous at $x=c$
If f(x) is continuous at is continuous at $x=c$.
My proof: for any $ϵ>0$, we need to show there exists $\delta >0$, such that whenever $|x-c|<\delta$, we have $|\frac{1}{f\left(x\right)}-\frac{1}{f\left(c\right)}|<ϵ$.
$|\frac{1}{f\left(x\right)}-\frac{1}{f\left(c\right)}|=|\frac{f\left(x\right)-f\left(c\right)}{f\left(x\right)f\left(c\right)}|$
$=|f\left(x\right)-f\left(c\right)|\frac{1}{|f\left(x\right)f\left(c\right)|}$
$=|f\left(x\right)-f\left(c\right)|\frac{1}{|f\left(x\right)|}\frac{1}{|f\left(c\right)|}$
Since f is continuous at c, thus $|f\left(x\right)-f\left(c\right)|<ϵ$. I need to find an upper bound of $|f\left(x\right)|$.
Because $|f\left(x\right)-f\left(c\right)|<ϵ$, we then have
$|f\left(c\right)|=|f\left(c\right)-f\left(x\right)+f\left(x\right)|\le |f\left(c\right)-f\left(x\right)|+|f\left(x\right)|$.
$<ϵ+|f\left(x\right)|$
Thus, we find $|f\left(x\right)|>|f\left(c\right)|-ϵ$. Equivalently, $\frac{1}{|f\left(x\right)|}<\frac{1}{|f\left(c\right)|-ϵ}$. (Not rigorous here, since $|f\left(c\right)|-ϵ$ could be negative, how to remedy?)
Finally, (ignore not rigorous part)