As part of my Real Analysis-I homework, we were given the following question:
Let $ f(x) = [x] $ i.e. the greatest integer function. Fix a point $ p \in \mathbb{R} $. Using either of the definitions of the limit of $ f(x) $ as $x$ approaches $p$ -- whichever is more convenient -- discuss whether or not $f(x)$ has a limit as $x$ approaches $p$, depending on $p$.
I decided to try and do it using the $ \epsilon-\delta $ definition (although we have also discussed the sequential definition in class). My proof is as below, but I'm not sure if it is rigorous enough (especially since I wasn't able to define a $ \delta $ in terms of $ \epsilon $.)
Consider $ x \in \mathbb{R} $, then we know that $ \exists \ a \in \mathbb{Z} $ such that $ a \le x \le (a+1) $, as $\mathbb{Z}$ is not bounded in $\mathbb{R}$, by the Archimedean Property.
Case 1: Take $ I = (a, a+1), \ p \in I$. Then $ \forall x \in I, [x] = a. $ Fix $ \epsilon > 0,$ choose $ \delta_1 > 0 $ such that $ 0 < |x-p| < \delta_1, \forall x \in I $
Then $ |f(x) - a| = |[x] - a| = |a - a| = 0 < \epsilon \ \forall x \in I, 0 < |x-p| < \delta_1 $
Thus, by definition, $ lim_{x \rightarrow p} = a $.
Case 2: Take $ I = (a - \frac{1}{2}, a+\frac{1}{2}), \ p = a \in I$.
Then $ \begin{equation} [x] = \begin{cases} a-1 & \text{if } x \in (a - \frac{1}{2}, a)\\ a & \text{if } x \in (a, a + \frac{1}{2}) \end{cases} \end{equation} $
Let us assume $f(x)$ has the limit $L$. Then fix $ \epsilon > 0,$ choose $ \delta_2 > 0 $ such that $ 0 < |x-p| < \delta_2, \forall x \in I $.
Now $ |f(x) - L| = |[x] - L| < \epsilon $
$ \implies |a - 1 - L| < \epsilon \ \forall x \in (a - \frac{1}{2}, a) $
$ and \ \ |a - L| < \epsilon \ \forall x \in (a, a + \frac{1}{2}) $
Now consider $ 1 = |a - L + L - a - 1| $
$ \le |a - L| + |a - 1 - L| $ by triangle inequality
$ < \epsilon + \epsilon = 2\epsilon $
Thus, we have $ 1 < 2\epsilon \implies \epsilon > \frac{1}{2} $ which is not true $ \forall \epsilon > 0 $
Hence, $ \exists \epsilon > 0 \ \text{such that} \ |f(x) - L| \ge \epsilon \ \forall \delta > 0, \text{such that} \ 0 < |x-p| < \delta, \text{thus} \ f(x) $ does not have the limit $L$ as $x \rightarrow p$.
So, we see that the limit exists only when p is not an integer.
I'm not entirely convinced by this proof that I have come up with, because (as mentioned above), I'm not able to concretely define a $ \delta $ in terms of $ \epsilon $.
Please help me solve the problem I'm facing above, and also please find any other holes in my proof! If you can help me with an alternate, more correct proof, then that would be great.
Notes: (1) I'm a first year undergraduate, so my knowledge is a little limited - sorry if I omitted something important somewhere, but I'm open to learning! (2) This was my first time using LaTeX/MathJax, so I'm sorry if I did something stupid :(