$\epsilon$–$\delta$ proof that $\lim_{x\to x_0}\frac{1}{x} = \frac{1}{x_0}$ for all $x_0\neq 0$; how to identify $\delta$?

Question

I would like to use the $\epsilon$–$\delta$ definition of the limit of a function to show that $$\lim_{x\to x_0} \frac{1}{x} = \frac{1}{x_0}$$ But I'm having trouble identifying a $\delta>0$ for arbitrary $\epsilon>0$ and $x_0\neq 0$ so that $$ 0<|x-x_0|<\delta \implies |\frac{1}{x} -\frac{1}{x_0}|<\epsilon$$ How can you find a $\delta$ that satisfies this condition?

Answer 1

Examine the $4$ cases of $|x-x_0|<\delta$ indivdually.

$$x<\delta+x_0,\;\;x_0>0$$

$$x<\delta+x_0,\;\;x_0<0$$

$$x<-\delta+x_0,\;\;x_0>0$$

$$x<-\delta+x_0,\;\;x_0<0$$

Now

$$\left|\frac{1}{x}-\frac{1}{x_0}\right|$$

$$=\left|\frac{x-x_0}{x_0(\delta+x_0)}\right|$$

$$<\left|\frac{x-x_0}{x_0^2}\right|$$

which gives the answer for this case, and you will need to go through each case separately.

Answer 2

Given $\ \varepsilon > 0.$

If $\varepsilon < \frac{1}{\vert x_0 \vert },$ then $\vert \varepsilon x_0 \vert<1,\ $ and so $$ x \in \left( \underbrace{\ \frac{x_0}{1 + \varepsilon x_0}\ }_{a}, \underbrace{\ \frac{x_0}{1- \varepsilon x_0}\ }_{b} \right) \implies \frac{1}{x} \in \left( \frac{1 - \varepsilon x_0}{x_0}, \frac{1+ \varepsilon x_0}{x_0} \right) = \left( \frac{1}{x_0} - \varepsilon, \frac{1}{x_0} + \varepsilon \right). $$

Else, if $\ \varepsilon \geq \frac{1}{\vert x_0 \vert }\ \left( > \frac{1}{2}\cdot \frac{1}{\vert x_0 \vert} \right),\ $ then $$ x \in \left( \underbrace{\ \frac{1}{ \frac{1}{x_0} + \frac{1}{2} \cdot \frac{1}{\vert x_0 \vert}\ } }_{c}, \underbrace{\ \frac{1}{ \frac{1}{x_0} - \frac{1}{2} \cdot \frac{1}{\vert x_0 \vert} }\ }_{d} \right) \implies \frac{1}{x} \in \left( \frac{1}{x_0} - \frac{1}{2} \cdot \frac{1}{\vert x_0 \vert}, \frac{1}{x_0} - \frac{1}{2} \cdot \frac{1}{\vert x_0 \vert} \right)$$

$$\subset \left( \frac{1}{x_0} - \varepsilon, \frac{1}{x_0} + \varepsilon \right). $$

So let $$\delta:= \begin{cases} \frac{b-a}{2}&\text{if}\, \varepsilon < \frac{1}{\vert x_0 \vert }\\ \frac{d-c}{2}&\text{if}\, \varepsilon \geq \frac{1}{\vert x_0 \vert }\\ \end{cases} $$