If $|x -a| < \min\{|a|/2,\, \varepsilon a^2 /2\}$ then... 
Suppose $a \neq 0$ and $\varepsilon >0$.  If $$|x-a| < \min \left \{ \frac{|a|}{2}, \frac{\varepsilon a^2 }{2}\right \},$$
then show that $x \neq 0$ and $\left |\frac{1}{x}-\frac{1}{a}\right| < \varepsilon$.

Proof:
Let $\delta = \min \left \{ \frac{|a|}{2}, \frac{\varepsilon a^2 }{2}\right \}$, where $\delta \leq \frac{|a|}{2}, \frac{\varepsilon a^2}{2}$ , $0< |x - a| < \delta$.
Then $|\frac{1}{x} - \frac{1}{a}| = |\frac{a-x}{ax}|$....
I was able to solve this when x and a are fixed positive. But the question only specifies that $x,a \neq 0$.

Case 1: $a>0, |a|\le\varepsilon a^2$
In this particular case:
$$\varepsilon a\ge1$$
$$|x-a|<\frac a2\implies$$
$$-\frac{a}{2}<x-a<\frac a2$$
$$\frac{a}{2}<x<\frac {3a}2 \quad \\ \text{(notice: all values are positive)}$$
$$\frac{2}{a}>\frac 1x>\frac {2}{3a}$$
$$\frac{2}{a}-\frac 1a>\frac 1x-\frac 1a>\frac {2}{3a}-\frac 1a$$
$$\frac{1}{a}>\frac 1x-\frac 1a>-\frac {1}{3a}$$
$$\left|\frac 1x-\frac 1a\right|<\frac 1a=\frac
\varepsilon{a\varepsilon}\le\varepsilon$$
Case 2: $a>0, |a|>\varepsilon a^2$
In this particular case: 
$$\varepsilon a<1$$
$$|x-a|<\frac {\varepsilon a^2}2\implies$$
$$-\frac {\varepsilon a^2}2<x-a<\frac {\varepsilon a^2}2$$
$$a-\frac {\varepsilon a^2}2<x<a+\frac {\varepsilon a^2}2\\ \text{(notice: all values are positive)}$$
$$\frac{2}{a(2-\varepsilon a)}-\frac 1a>\frac 1x-\frac 1a >\frac{2}{a(2+\varepsilon a)}-\frac 1a$$
$$\frac{\varepsilon}{2-\varepsilon a} >  \frac 1x-\frac 1a  > -\frac{\varepsilon}{2+\varepsilon a}$$
$$\frac{\varepsilon}{2-\varepsilon a} >  \left|\frac 1x-\frac 1a \right|\tag{1}$$
But $\varepsilon a<1 \implies 2-\varepsilon a>1$ or:
$$\frac1{2-\varepsilon a}<1$$ 
$$\frac\varepsilon{2-\varepsilon a}<\varepsilon\tag{2}$$
Combine (1) and (2) and you get:
$$\varepsilon >  \left|\frac 1x-\frac 1a \right|$$ 
Case 3: $a<0, |a|\le\varepsilon a^2$
Intorduce $b=-a$ so that $b$ is positive. In this particular case:
$$\varepsilon b>=1$$ 
We have to prove the following implications:
$$|x+b|<\frac b2\implies$$
$$-\frac b2<x+b<\frac b2$$
$$-b-\frac b2<x<\frac b2 - b$$
$$-\frac {3b}2<x<-\frac b2\\ \text{(notice: all numbers are negative)}$$
$$-\frac {2}{3b}+\frac 1b>\frac 1x + \frac 1b>-\frac 2b+\frac 1b$$
$$-\frac {2}{3b}+\frac 1b>\frac 1x + \frac 1b>-\frac 2b+\frac 1b$$
$$\frac {1}{3b}>\frac 1x + \frac 1b>-\frac 1b$$
$$\left|\frac 1x + \frac 1b\right|<\frac 1b=\frac {\varepsilon}{\varepsilon b}\le\varepsilon$$
$$\left|\frac 1x - \frac 1a\right|<\varepsilon$$
Case 4: $a<0, |a|>\varepsilon a^2$
Introduce $b=-a$ so that $b$ is positive. In this particular case:
$$\varepsilon b<1$$ 
We have to prove the following implications:
$$|x+b|<\frac {\varepsilon b^2}2\implies$$
$$-\frac {\varepsilon b^2}2<x+b<\frac {\varepsilon b^2}2$$
$$-b-\frac {\varepsilon b^2}2<x<-b+\frac {\varepsilon b^2}2\\ \text{(notice: all values are negative)}$$
$$-\frac{2}{b(2+\varepsilon b)}+\frac 1b>\frac 1x+\frac 1b >\frac{2}{b(-2+\varepsilon b)}+\frac 1b$$
$$\frac{\varepsilon}{2+\varepsilon b} >  \frac 1x+\frac 1b  > -\frac{\varepsilon}{2-\varepsilon b}$$
$$\left|\frac 1x+\frac 1b \right|<\frac{\varepsilon}{2-\varepsilon b}\tag{3}$$
But $\varepsilon b<1 \implies 2-\varepsilon b>1$ or:
$$\frac1{2-\varepsilon b}<1$$ 
$$\frac\varepsilon{2-\varepsilon b}<\varepsilon\tag{4}$$
Combine (3) and (4) and you get:
$$\left|\frac 1x+\frac 1b \right|=\left|\frac 1x-\frac 1a \right|<\varepsilon$$ 
