I have to prove that the function $f\colon [0,\infty)\to R$, defined by $f(x)=\lbrack\!\lbrack \sqrt{x}\rbrack\!\rbrack$ $\forall$ $x\in [0,\infty)$, is continuous at $c=2$. The function $f(x)=\lbrack\!\lbrack \sqrt{x}\rbrack\!\rbrack$ takes a number and rounds it down to the closest integer if anyone's unfamiliar with the notation. So $f(x)=\lbrack\!\lbrack {1.999999999}\rbrack\!\rbrack$ would be $1$. This is what I originally put as an idea:
Let us look at the values from $f:[1,2)$ and $f:(2,3]$. We see by the greatest integer function that $f:[1,2)$ outputs all $1$'s and all values produced by $f:(1,2]$ are all $1$'s as well. In fact, $f:[1,4)$ gives all $1$'s. The limit as $x$ approaches $2$ of $f:[1,2)$ is $1$ and the limit as $x$ approaches $2$ of $f:(2,3]$ is $1$, which are equal and thus the function is continuous at $c=2$.
The feedback I received on my problem was that I had the right idea but to rather restrict my attention to $V_1(2)$ for building an inequality for an epsilon-delta proof. I would really like some help learning how to correctly set up an epsilon-delta proof in this case as I am used to using these proofs only for proving limits and not necessarily proving continuity. Thanks in advance!
This is a "simple" delta-epsilon proof as the delta needn't be dependent on the epsilon.
For x = 2 simply let $\delta = 1$. Then $|2 - y| < 1 \implies y \in (1,3)$ and $1< \sqrt{y} < 2$ so $f(y) = 1$ and $f(2) = 1$ so $|f(2) - f(x)| = |1 - 1| = 0$.
Therefore
$|2 -y| < \delta \implies |f(2) - f(y)| < \epsilon$ for all $\epsilon > 0$. Hence $f$ is continuous at 2.
==============
In general you use delta epsilon proves for continuity the same way you use it for limits.
Definition: $f$ is continuous at x if for any $\epsilon > 0$ there exists a $\delta$ such that for any y such that $|x - y| < \delta$ it follows that $|f(x) - f(y)| < \epsilon$.
To prove this you usually start with setting
$|f(x) - f(y)| < \epsilon$ and trying to manipulate this to get $|x - y| < \text{some value based on}\, \epsilon = \delta$.
So in this case
$|f(2) - f(y)| < \epsilon$ => $|1 - f(y)| < \epsilon$. But as f(y) is an integer and as $\epsilon$ could/should be less than 1 for this to be true this means $|1 - f(y)| = 0 \iff f(y) = 1$.
So what are the possible values that force $f(y) = 1$. That would mean $1 < y < 4$. So how close to 2 does y have to be? Well, as long as $|2 -y| < 1$ this will follow. So $\delta$ can be 1.