Proving $\lim_{x \rightarrow a} \frac{1}{x} = \frac{1}{a}$ by use of limit definition

Question

$$ \frac{-|x-a|}{a|x|} < \varepsilon $$

Then i did was assume $\delta = a/2$ and got the values where $x \in (a/2,3a/2)$. Idk what to do next and im sorry idk how to format.

Answer 1

You are on the right track. Let's assume that $x>0$ and $a>0$.

We first take $a/2<x<3a/2$, which means that $|x-a|<a/2$.

Next, let $\epsilon>0$ be given. Then, we can write

$$\begin{align} \left|\frac1x-\frac1a\right|&=\frac{|x-a|}{ax}\\\\ &<\frac{|x-a|}{a(a/2)}\\\\ &<\epsilon \end{align}$$

whenever $|x-a|<\min\left(a/2,\frac12 a^2\epsilon\right)$.

The case for $x<0$, $a<0$ follows an analogous line and is left as an exercise.