$ε$-$δ$ proof for continuity of $\frac1x$

Question

So $f(x) = \dfrac{1}{x}$ for $x \ne 0 $ and $f(x) = 20$ if $x = 0$, and you need to prove that it is continuous or discontinuous at $x = 0$.

So we want to show that $\lvert f(x)-f(0) \rvert < ε$. I got so far as to show $\left\lvert\dfrac{1}x - 20 \right\rvert$ and that $x < \delta$. Now I know that the function is discontinuous, so is it okay to say that if $$ε = \frac{1}{\delta} - 20$$ then $$ \frac{1}x-20 < \frac{1}\delta - 20$$ $$ \frac{1}x < \frac{1}\delta$$ $$\delta < x$$

Which cannot happen so it is discontinuous?

Answer 1

As noted in the comments, you want to negate the definition of continuity.

So, we are in search of some $\epsilon>0$, let's take $\epsilon=1$, with the property that for any $\delta>0$, we have $|x|<\delta$ and $|1/x_*-20|\geq 1$ for some $|x_*|<\delta$.

I'll tackle the limit from the right, and leave the case when $x<0$ to you.

Fix an arbitrary $\delta>0$. If $0<x<\delta$ we know $\frac{1}{x}>\frac1\delta$, so as long as $$ \frac1\delta\geq 21 $$ we have $$ \frac{1}{x}\geq21\implies \frac{1}{x}-20\geq1 $$ and $$ \left|\frac1x-20 \right|>1 $$ for all of the $x$ with $x<\frac{1}{21}$ in our delta ball.

Answer 2

This assumption $$ ε = \frac{1}{\delta } - 20 $$does not make your $\epsilon$ positive for $ \delta =1$ or many other numbers. How do you address that ?