I need to show that if $|y-y_0|<\text{min}(\frac{|y_0|}{2},\frac{\epsilon|y_0|^2}{2})$ and $y\neq 0$ and $y_0\neq 0$ are true, then the following inequality is also true: $$ \left\vert \frac{1}{y}-\frac{1}{y_0} \right\vert < \epsilon.$$
I tried to check the information given by the inequalities $|y-y_0|<\frac{|y_0|}{2}$ and $|y-y_0|<\frac{\epsilon|y_0|^2}{2}$. Then I arrived from those inequalities to these ones: $|\frac{y}{y_0}-1|<\frac{1}{2}$ and $|\frac{1}{y} \cdot \frac{y^2}{y_0^2}-\frac{1}{y_0}|<\frac{\epsilon}{2}$.
I multiplied the inequality $|y-y_0|<\frac{\epsilon|y_0|^2}{2}$ by $\frac{1}{|y y_0|}$ , because $\frac{1}{|yy_0|}>0$. It follows that $|\frac{1}{y_0}-\frac{1}{y}|<\frac{\epsilon|y_0|}{2|y|}$. Because $|\frac{1}{y_0}-\frac{1}{y}|=|\frac{1}{y}-\frac{1}{y_0}|$ is true for the absolute value function, we get $|\frac{1}{y}-\frac{1}{y_0}|<\frac{\epsilon|y_0|}{2|y|}$...
I am still trying to figure out, how I can prove $ \left\vert \frac{1}{y}-\frac{1}{y_0} \right\vert < \epsilon.$
Any help is appreciated. Thanks in advance.
Regards,
Ahmed Hossam
Case 1:
$$\text{min}(\frac{|y_0|}{2},\frac{\epsilon|y_0|^2}{2})=\frac{|y_0|}{2}$$
$$\frac{|y_0|}{2}\le\frac{\epsilon|y_0|^2}{2}$$
$$\frac1{|y_0|}\le\epsilon\tag{1}$$
In this particular case:
$$|y-y_0|\lt \frac{|y_0|}{2}$$
$$-\frac{|y_0|}{2}<y-y_0\lt \frac{|y_0|}{2}$$
$$y_0-\frac{|y_0|}{2}<y< y_0+\frac{|y_0|}{2}\tag{2}$$
If $y_0>0$, (2) becomes:
$$\frac{y_0}{2}<y< \frac{3y_0}{2}$$
$$\frac{2}{y_0}>\frac1y> \frac{2}{3y_0}$$
$$\frac{2}{y_0}-\frac{1}{y_0}>\frac1y - \frac{1}{y_0}> \frac{2}{3y_0}-\frac{1}{y_0}$$
$$\frac{1}{y_0}>\frac1y - \frac{1}{y_0}> -\frac{1}{3y_0}$$
$$\frac{1}{y_0}>|\frac1y - \frac{1}{y_0}|\tag{3}$$
Because of (1) and (3) we have:
$$\epsilon>|\frac1y - \frac{1}{y_0}|$$
If $y_0<0$, (2) becomes:
$$y_0+\frac{y_0}{2}<y< y_0-\frac{y_0}{2}$$
$$\frac{3y_0}{2}<y< \frac{y_0}{2}$$
Notice that $y$ is squeezed between negative values so we can write:
$$\frac2{3y_0}>\frac1y> \frac2{y_0}$$
$$\frac2{3y_0}-\frac{1}{y_0}>\frac1y-\frac1{y_0}> \frac2{y_0}-\frac1{y_0}$$
$$-\frac{1}{3y_0}>\frac1y-\frac1{y_0}> \frac1{y_0}$$
$$|\frac1y-\frac1{y_0}|< \frac1{|y_0|}\tag{4}$$
Because of (1) and (4) we have:
$$|\frac1y-\frac1{y_0}|< \epsilon$$
Case 2:
$$\text{min}(\frac{|y_0|}{2},\frac{\epsilon|y_0|^2}{2})=\frac{\epsilon|y_0|^2}{2}$$
$$\frac{|y_0|}{2}\ge\frac{\epsilon|y_0|^2}{2})$$
$$1\ge\epsilon|y_0|\tag{5}$$
In this particular case:
$$|y-y_0|\lt \frac{\epsilon|y_0|^2}{2}$$
$$-\frac{\epsilon|y_0|^2}{2}\lt y-y_0\lt \frac{\epsilon|y_0|^2}{2}$$
$$y_0-\frac{\epsilon|y_0|^2}{2}\lt y \lt y_0+\frac{\epsilon|y_0|^2}{2}\tag{6}$$
For $y_0>0$, (6) becomes:
$$y_0-\frac{\epsilon y_0^2}{2}\lt y \lt y_0+\frac{\epsilon y_0^2}{2}$$
Notice that because of (5) $y$ is squeezed between positive values so we can write:
$$\frac{1}{y_0-\frac{\epsilon y_0^2}{2}}\gt \frac1y \gt \frac1{y_0+\frac{\epsilon y_0^2}{2}}$$
$$\frac{1}{y_0-\frac{\epsilon y_0^2}{2}}-\frac 1{y_0} \gt \frac1y-\frac 1{y_0} \gt \frac1{y_0+\frac{\epsilon y_0^2}{2}}-\frac 1{y_0}$$
This evolves into:
$$\frac{\epsilon}{2-\epsilon y_0}>\frac1y-\frac 1{y_0}>-\frac{\epsilon}{2+\epsilon y_0}\tag{7}$$
Notice that because of (5):
$$2-\epsilon y_0 \ge 1 \implies \epsilon \ge \frac{\epsilon}{2-\epsilon y_0}$$
$$2+\epsilon y_0 \ge 2 \implies -\frac{\epsilon}{2+\epsilon y_0}>-\frac{\epsilon}{2}$$
With this in mind, (7) now becomes:
$$\epsilon >\frac1y-\frac 1{y_0}>-\frac{\epsilon}{2}$$
$$\epsilon >|\frac1y-\frac 1{y_0}|$$
I'm leaving the last case ($y_0<0$) to you as an exercise.