Given $x \in \left ( \frac{1}{10}, \frac{3}{10} \right)$. Find $\delta > 0$ so that $$0< \left| x- \frac{1}{5} \right| < \delta \Rightarrow \left| \frac{1}{x} - 5\right| < \frac{1}{10}.$$ How can I answer that question?
My Idea
Let's do a little observation: $$\left| \frac{1}{x} - 5\right| < \frac{1}{10} \Longleftrightarrow -\frac{1}{10} < \frac{1}{x}-5 <\frac{1}{10} \Longleftrightarrow \frac{49}{10}< \frac{1}{x}<\frac{51}{10}.$$ We get $$\frac{10}{51} < x < \frac{10}{49} \Longleftrightarrow \frac{10}{51}-\frac{1}{5}< x - \frac{1}{5} < \frac{10}{49} - \frac{1}{5}.$$
As a consequence $$\frac{-1}{51 \cdot 5} < x - \frac{1}{5}< \frac{1}{49 \cdot 5} \Longleftrightarrow \frac{-1}{255} < x - \frac{1}{5} < \frac{1}{255}< \frac{1}{245}.$$
We obtain $$\left| x - \frac{1}{5}\right| < \frac{1}{255}.$$
Choose $\delta = \frac{1}{255}$, so that the implication above is true (?)
How can I fix this problem? Is my idea correct?.
Your idea is correct. As you may directly verify, you have the implication $$0< \left| x- \frac{1}{5} \right| < \frac1{255} \;\;\Rightarrow\;\; \left| \frac{1}{x} - 5\right| < \frac{1}{10}.$$ Sometimes I prefer to avoid double inequalities (i.e. $A<x<B$), since they can be difficult to look at if there is a longer sequence of steps. Here we can get a hint by looking at the graph of $f(x)=1/x.$ For any $\delta$-neighborhood of $1/5$ consisting of strictly positive inputs $x$, the graph will always be steeper throughout the left-hand half of the interval, compared with all the right-hand half. So we suspect the tougher requirement on $\delta$ will be found for $x\in(-\delta+1/5,\;1/5)$.
Indeed, if $0<x<1/5,$ then $\frac1x-5>0$, and $|\frac1x-5|=\frac1x-5.$ Now we require $$\frac1x-5 \;<\; \frac1{10}.$$
But you can readily check that this occurs iff: $$\frac15-x \;<\;\frac15 - \frac{10}{51} \;\quad\left(=\;\frac1{255}\right).$$
On the other hand, you can repeat the above for the right-hand half (where $f$ is comparatively flat) and obtain
$$5-\frac1x \;<\; \frac1{10} \;\;\iff\;\; x-\frac15 \;<\;\frac{10}{49} - \frac{1}{5} \;\quad\left(=\;\frac1{245}\right).$$