Q:

Proving continuity using the epsilon definition,a specific case

So, I was arguing with my friend about who is correct, I will not say what I said so in the case that I was wrong you will not know it :)
Anyway, the question is:
Prove that $f(x)= \frac1x$ is continuous at $x=2$ using the epsilon definition.
The function is defined at $x=2$ , $f(2)=\frac12$
Take $\epsilon > 0$
We need:
$$|f(x) - f(2)| < \epsilon\\
|\frac1x - \frac12| < \epsilon\\
-\epsilon < \frac1x - \frac12 < \epsilon\\
\frac12 - \epsilon < \frac1x < \epsilon + \frac12$$
Since $x=2$ and $1/2 \pm \epsilon$ is a neighborhood of $x=2$ the function is continuous at the point.
Is this fine or is there something missing ?

A:

The continuity of function $f$ at some point $x$ requires for any given $\varepsilon>0$, there exists $\delta>0$ such that for any $y$ such that $|y-x|<\delta$, we have $|f(y)-f(x)|<\varepsilon$.
So actually you are on the road, yet not reach the final. What you need is to find a $\delta$ so that the continuity condition holds. Could you keep on going now?

