I am still getting confused with showing discontinuity of functions, here is my attempt at a question , if someone could check to see if I am going about it in the correct way.
$f: \mathbb{R} \rightarrow \mathbb{R}$ given by $f(x)=[x]$ is discontinuous at 0.
My attempt;
I decided against using the epsilon-delta definition and decided to go with the limit approach.
We want to show that $$\lim_{x \to 0}[x] \neq f(0)=0$$
Now suppose it is, then $\forall \epsilon>0 \exists \delta >0$ s.t $|f(x)-f(c)|<\epsilon $whenever $|x-c|<\delta$.
Take $\epsilon =\dfrac{1}{2}$ ; for all $x<0$ $f(x)<0$ so if $f(x)<0$ using our chosen delta we want $|f(x)-f(0)|=|f(x)|<\dfrac{1}{2}$ but this is impossible , as there exists no $\delta$ that satisfies $|f(x)|<\dfrac{1}{2}$ whenever $|x-c|<\delta$
Is this correct??
You over complicate things. Simply take the left hand limit and the right hamd limit seperately. You will find the left hand limit to be -1 and the right hand limit to be 0. Therefore, there is jump discontinuity at x=0.