Find a delta for a specific epsilon. Let $f: [1, \infty) \rightarrow \mathbb{R}$, $f(x) = \frac{x-[x]}{[x]}$ where [x] is the largest integer $n$ with $n\le x$.
The question asks to find $\delta > 0$ such that if $|x-3|< \delta$ then $|f(x)-f(3)|< \frac{1}{2}$.
When I tried to solve this, I found out that $f(x)$ is not continuous at $3$ and that $f(x)$ approaches $\frac{1}{2}$ as $x$ approaches $3$, so I cannot find any $\delta$ that is suitable, and I am not sure what to do.
Any help would be appreciated.
 A: You could use $\delta = 1$ since  from $2$ up to just to the left of $3$ the function goes up from just above $0$ to, but not all the way up to $+1/2$, while exactly at $3$ it is $0,$ and from $3$ up to just before $4$ it goes only up to just below $1/3$.
You're right about it's not being continuous at $3$, that would require that for every $\varepsilon$ there is a $\delta$ etc.
A: *

*For $x \in (2,3)$ we have $$f(x) = \frac{x-2}{2}.$$ Since $f(3)=0$, $$|f(x)-f(3)| = |f(x)|< \frac{1}{2}$$ is satisfied if $$\left| \frac{x-2}{2} \right| < \frac{1}{2},$$ i.e. if $|x-2| <1$. Thus, $|f(x)-f(3)|< \frac{1}{2}$ for all $x \in (2,3)$.

*Let $x \in (3,4)$. Then $$f(x) = \frac{x-3}{3}; $$ hence$$|f(x)-f(3)|< \frac{1}{2}$$ if $$\left| \frac{x-3}{3} \right| < \frac{1}{2},$$ i.e. if $|x-3|< \frac{3}{2}$. This means that $|f(x)-f(3)| < \frac{1}{2}$ holds for all $x \in (3,4)$.


Consequently, we can choose any $\delta \in (0,1)$.
A: $f(3) = 0$. So you need to find a $\delta$ such that in the open interval $(3-\delta, 3+\delta)$, $f(x) < \frac{1}{2}$.
As you observed, the limit  of $f(x)$ as $x$ approaches $3$ from below is $1\over 2$. But that is no problem since in the open interval $(2,3)$, $f(x) = \frac{x-2}{2} < \frac{3-2}{2} = \frac{1}{2}$.  And for all $x \geq 3$, $f(x) < \frac{1}{3} < \frac{1}{2}$.
So for any $\delta \leq 1$, $x \in (3-\delta, 3+\delta) \rightarrow f(x) < \frac{1}{2}$.
Since  $x \in (\frac{3}{2},2) \rightarrow  f(x) > \frac{1}{2}$, the largest $\delta$ you could use that satisfies the requirement is in fact $\delta = 1$.  
