Show that $f(x)=1/x$ is a continuous function for any $x\neq 0$

Question

Following a reference from “Elementos de Topología general” by Angel Tamariz and Fidel Casarrubias.

Let be $a\in\mathbb{R}\setminus\{0\}$ and we choose $\epsilon>0$. We define $\delta=\min\left\{\frac{|a|}2,\frac{\epsilon|a|^2}2\right\}$ and so we observe that if $y\in B(a,\delta)$, then $$ \left|\frac{1}y-\frac{1}a\right|=\frac{|y-a|}{|y||a|}\le\frac{2}{|a|^2}|y-a|<\frac{2}{|a|^2}\frac{|a|^2\epsilon}2=\epsilon $$

and so we conclude that $\frac{1}x$ is continuous.

Unfortunately I don't understand why $\frac{|y-a|}{|y||a|}\le\frac{2}{|a|^2}|y-a|$, that is $\frac{1}{|y|}\le\frac{2}{|a|}$. In fact we have $$ |y-a|<\delta\le\frac{|a|}2\Rightarrow-2|a|<a-\frac{|a|}2<y<a+\frac{|a|}2<2|a|\Rightarrow|y|<2|a|\Rightarrow\frac{1}{|a|}<\frac{2}y $$ that is in contradiction with $\frac{1}{|y|}\le\frac{2}{|a|}$. Could someone help me, please?

Answer 1

Since $\lvert y-a\rvert<\frac{\lvert a\rvert}2$, we have$$\lvert y\rvert\geqslant\bigl\lvert\lvert a\rvert-\lvert y-a\rvert\bigr\rvert=\lvert a\rvert-\lvert y-a\rvert>\frac{\lvert a\rvert}2.$$So$$\frac{\lvert y-a\rvert}{\lvert y\rvert\lvert a\rvert}<\frac{\lvert y-a\rvert}{\lvert a\rvert}\times\frac2{\lvert a\rvert}=\frac2{\lvert a\rvert^2}\lvert y-a\rvert.$$

Answer 2

$$ |a| = |y|\leq |a-y| < \delta < \frac{|a|}{2}$$

Thus

$$ |y| > \frac{|a|}{2}$$