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

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 $$|y-a|<\delta\le\frac{|a|}2\Rightarrow-2|a| that is in contradiction with $$\frac{1}{|y|}\le\frac{2}{|a|}$$. Could someone help me, please?

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$$\frac{\lvert y-a\rvert}{\lvert y\rvert\lvert a\rvert}<\frac{\lvert y-a\rvert}{\lvert a\rvert}\times\frac2{\lvert a\rvert}=\frac2{\lvert a\rvert^2}\lvert y-a\rvert.$$

• Sorry, could you prove that $\lvert y\rvert\geqslant\bigl\lvert\lvert a\rvert-\lvert y-a\rvert\bigr\rvert=\lvert a\rvert-\lvert y-a\rvert>\frac{\lvert a\rvert}2$? Unfortunately it seems that it is difficult to me. Mar 4 '20 at 23:21
• Note that$$\lvert a\rvert=\lvert a-y+y\rvert\leqslant\lvert a-y\rvert+\lvert y\rvert$$and that therefore $\lvert y\rvert\geqslant\lvert a\rvert-\lvert y-a\rvert$. Now, use the fact that $\lvert y-a\rvert<\frac{\lvert a\rvert}2$. Mar 4 '20 at 23:24
• Sorry, but I didn't succeed: I only see that $|y-a|<\frac{|a|}2<|a|\Rightarrow 0<|a|-|y-a|$. Could you explain, please? Mar 4 '20 at 23:35
• Since $\lvert y-a\rvert<\frac{\lvert a\rvert}2$, then $\lvert a\rvert-\lvert y-a\rvert>\lvert a\rvert-\frac{\lvert a\rvert}2=\frac{\lvert a\rvert}2.$ Mar 4 '20 at 23:38
• Okay, now it is clear. Thanks! Mar 4 '20 at 23:42

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

Thus

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