Proving $\lim\limits_{x \to 3/2}{[x]}=1$

Question

$[x]$ is the largest integer less than or equal to $x$

so $[\frac{3}{2}]=1$

Using epsilon and delta, how can I show that this limit exists?

Answer 1

For $\forall x\in [1,2): [x]=1$. Therefore $\forall \epsilon>0$ always exists $\delta>0$, for example $\delta=0.3$ such that $0=|[x]-1|<\epsilon,\forall x: |x-\frac{3}{2}|<\delta=0.3$ which means for all $x\in(\frac{3}{2}-\delta,\frac{3}{2}+\delta)=(1.2,1.8)$