Proof that $1/f(x)$ is bounded

Question

I wish to prove the following theorem.

Statement: If $$\lim_{x\to \ a} f(x)=b\neq0,$$ then the function $$\frac{1}{f(x)}$$ is bounded for $x\rightarrow a.$

Attempt

From the definition of a limit, $$\| f(x)-b \|<\varepsilon.$$

Taking the absolute value from both sides, remembering that $\epsilon>0$ yields $$\|f(x)-b\|<\varepsilon.$$

Since in general $|w-z|\ge |w|-|z|$, then$$\|f(x)-b\|\ge ||f(x)| - |b||$$ and consequently $$||f(x)|-|b||<\varepsilon\iff-\varepsilon<|f(x)|-|b|<\varepsilon.$$

Add $|b|$ to the inequality.

$$|b|-\varepsilon<|f(x)|<\varepsilon+|b|$$ This is where I get stuck, however my book on analysis does not. According to the text, this result implies

$$\frac{1}{|b|-\varepsilon} \operatorname*{>}^\text{?} \frac{1}{|f(x)|} > \frac{1}{\varepsilon+|b|}.$$

Question

The part with the question mark is the one I am having trouble with. More specifically, how can one be sure that $$|b|-\varepsilon>0\ \text{ or } \ |b| > \varepsilon\text{?}$$

I am even more confused because

$$\|f(x)-b\| \ge ||f(x)|-|b|| > |-|b||=|b|$$

to me shows the opposite, i.e., $$|b|<\varepsilon.$$

What am I missing?

Answer 1

Remember the limit definition - it states that $|f(x) - b| <\varepsilon$ for all $\varepsilon > 0$. Because $b \neq 0$ you can choose an $\varepsilon$ such that $|b|>\varepsilon > 0$.

Also the statement $||f(x)| - b| > |-|b||$ is wrong - take $f(x) = 2$, $b = 3$ for example.

Answer 2

We have: $\displaystyle \lim_{x \to a} \dfrac{1}{f(x)} = \dfrac{1}{b}\implies $ take $\epsilon = 1$, then $\exists \delta > 0$ such that if $0 < |x-a| < \delta$ then $\left|\dfrac{1}{f(x)} - \dfrac{1}{b}\right|< 1\implies \left|\dfrac{1}{f(x)}\right|-\left|\dfrac{1}{b}\right| < \left|\dfrac{1}{f(x)}-\dfrac{1}{b}\right| < 1\implies \left|\dfrac{1}{f(x)}\right| < \dfrac{1+|b|}{|b|}$, which shows the function $g(x) = \dfrac{1}{f(x)}$ is bounded as $x \to a$.