The proof that if f(x) is continuous at x=c, then 1/f(x) is continuous at x=c

If f(x) is continuous at x=c, $f(x) \neq 0$ then 1/f(x) is continuous at x=c.

My proof: for any $\epsilon >0$, we need to show there exists $\delta >0,$ such that whenever $|x-c|< \delta$, we have $|\frac{1}{f(x)} - \frac{1}{f(c)}| < \epsilon$.

$$|\frac{1}{f(x)} - \frac{1}{f(c)}|= |\frac{f(x)-f(c)}{f(x)f(c)}|$$

$$=|f(x)-f(c)| \frac{1}{|f(x)f(c)|}$$

$$=|f(x)-f(c)| \frac{1}{|f(x)|} \frac{1}{|f(c)|}$$

Since f is continuous at c, thus $|f(x)-f(c)| < \epsilon$. I need to find an upper bound of $\frac{1}{|f(x)|}$, i.e., need to find a lower bound of $|f(x)|$.

Because $|f(x)-f(c)| < \epsilon$, we then have $|f(c)|= |f(c)-f(x)+f(x)| \leq |f(c)-f(x)|+|f(x)| < \epsilon +|f(x)|$. Thus, we find $|f(x)| > |f(c)| - \epsilon$. Equivalently, $\frac{1}{|f(x)|} < \frac{1}{|f(c)| - \epsilon}$. (Not rigorous here, since $|f(c)| - \epsilon$ could be negative, how to remedy?)

Finally, (ignore not rigorous part)

$$|\frac{1}{f(x)} - \frac{1}{f(c)}| < \frac{\epsilon}{|f(c)| - \epsilon} \frac{1}{|f(c)|}$$

Let $\epsilon$ be arbitrarily small, i.e., $\epsilon \to 0^+$ then $|f(c)| - \epsilon \to f(c)$, then

$$|\frac{1}{f(x)} - \frac{1}{f(c)}| < \frac{\epsilon}{|f(c)| ^2} $$

Hence, if at the beginning, let$|f(x)-f(c)| < \epsilon |f(c)| ^2$, then we can find that $$|\frac{1}{f(x)} - \frac{1}{f(c)}| < \epsilon $$.

The several last lines is not elegant. Maybe some small mistakes. How to remedy?

You can start by taking $\delta_0$ such that$$|x-c|<\delta_0\implies\bigl|f(x)-f(c)\bigr|<\frac{|f(c)|}2.$$Therefore$$|x-c|<\delta_0\implies\bigl|f(x)\bigr|>\frac{|f(c)|}2$$and so$$|x-c|<\delta_0\implies\left|\frac1{f(x)}-\frac1{f(c)}\right|<2\frac{|f(x)-f(c)|}{f^2(c)}.$$So, if you take $\delta_1>0$ such that$$|x-c|<\delta_1\implies\bigl|f(x)-f(c)\bigr|<\frac{\varepsilon f^2(c)}2,$$then$$|x-c|<\min\{\delta_0,\delta_1\}\implies\left|\frac1{f(x)}-\frac1{f(c)}\right|<\varepsilon.$$

Suppose we want to find a lower bound of $f(x)$ in a neighbourhood of $a$ where $f(a)\neq 0$. Since $f$ is continuous we can choose $|f(x)-f(a)|<\varepsilon$ in a neighbourhood of $a$. The trick here is choosing $\varepsilon=|f(a)|/2>0$. We then get $|f(x)-f(a)|<|f(a)|/2$. We also have $|f(a)|-|f(x)|\leq|f(x)-f(a)|$ by the triangle inequality, and we get $|f(a)|-|f(x)|<|f(a)|/2$. It then follows that $|f(a)|/2<|f(x)|$.