Question: Consider the function $f(x)=[x]$ on the interval $[0,2]$ where $[x]$ denotes the largest integer less than or equal to x.

Is this function continuous? I cant find a reason for it not to be, although im not sure.

Answer: 
If you're familiar with the properties of continuous functions, another way to see it is that $f^{-1}(1)=[0,1)$, which is not a closed subset of $[0,2]$.
**Hint:** Proof the negation of 
$$
(\forall \epsilon >0)(\exists \delta >0) \Big( (\forall x)(|x-1|<\delta \implies |f(x)-f(1)|<\epsilon)\Big),
$$
that is
$$
(\exists \epsilon >0)(\forall \delta >0) \Big( (\exists x)(|x-1|<\delta \mbox{ and } |f(x)-f(1)|>\epsilon)\Big)
$$