Find the largest possible delta-neighborhood to a given epsilon.

Question

Problem:

$ \lim_{x \to \pi} [[x]] = 3 $, where $\epsilon = 0.01$

([[x]] denotes the greatest integer less than or equal to x)

My thoughts:

Given a $\epsilon = 0.01$, $ \exists \delta > 0 $ such that $ 0< |x - \pi| < \delta \Rightarrow |[[x]] - 3| < \epsilon$. The Definition of $[[x]]$ gives $x - 4 < [[x]] - 3 < x -3 $.

And I am stuck from here as I don't know how to connect $ [[x]] - 3$ and $|x - \pi| $.

I appreciate your help! Thanks!

Answer 1

Intuitively, when $x$ is close to $\pi, [[x]]$ is guaranteed to be $3$. You just need to make sure $3 \le x \lt 4$ to have $[[x]]=3$