Is this function well-defined for all $\epsilon \in [0,1]$? Let
$$f(\epsilon)=\lim_{n \rightarrow \infty}\frac{\sum_{i=1}^n\left|\{k\le i^\epsilon: k |i\} \right|}{\sum_{i=1}^n\left|\{k |i\} \right|}$$
Where $| \cdot |$ denotes cardinality of a set.
Obviously we have $f(0)=0$, $f(0.5)=0.5$ and $f(1)=1$. But is this function well-defined for all $\epsilon \in [0, 1]$?
Edit: it also seems like $f(1-\epsilon)=1-f(\epsilon)$ for $\epsilon \in [0,1]$ such that $f(\epsilon)$ is well-defined.
 A: The limit always exists, and $f(\epsilon)=\epsilon$.
Call the numerator
$$ A(\epsilon,n) = \sum_{i=1}^n \Big|\{ k \in \mathbb{N} : 1 \leq k \leq i^\epsilon \land k \mid i \}\Big| $$
If $\delta: \mathbb{N} \times \mathbb{N} \to \mathbb{R}$ is the divisibility characteristic
$$ \delta(k,n) = \begin{cases} 1 & k \mid n \\ 0 & k \not\mid n \end{cases} $$
then
$$ A(\epsilon,n) = \sum_{i=1}^n \sum_{k=1}^{\lfloor i^\epsilon \rfloor} \delta(k,i) = \sum_{k=1}^{\lfloor n^\epsilon \rfloor} \sum_{i=\lceil k^{1/\epsilon}\rceil}^n \delta(k,i) $$
$$ A(\epsilon,n) = \sum_{k=1}^{\lfloor n^\epsilon \rfloor} \left( \left\lfloor\frac{n}{k}\right\rfloor - \left\lfloor \frac{\lceil k^{1/\epsilon} \rceil-1}{k}\right\rfloor \right) $$
Using the fact that if $x \leq y$ and $z>0$
$$ \left\lfloor \frac{y-x}{z} \right\rfloor \leq \left\lfloor \frac{y}{z} \right\rfloor - \left\lfloor \frac{x}{z} \right\rfloor \leq \left\lceil \frac{y-x}{z} \right\rceil, $$
$$ \begin{align*}
\sum_{k=1}^{\lfloor n^\epsilon \rfloor} \left\lfloor \frac{n-\lceil k^{1/\epsilon} \rceil+1}{k} \right\rfloor &\leq A(\epsilon,n) \leq \sum_{k=1}^{\lfloor n^\epsilon \rfloor} \left\lceil\frac{n-\lceil k^{1/\epsilon} \rceil+1}{k}\right\rceil \\
\sum_{k=1}^{\lfloor n^\epsilon \rfloor} \frac{n-k^{1/\epsilon}-k}{k} &\leq A(\epsilon,n) \leq \sum_{k=1}^{\lfloor n^\epsilon \rfloor} \frac{n-k^{1/\epsilon}+k+1}{k} \\
-\lfloor n^\epsilon \rfloor + \int_1^{n^\epsilon} \frac{n-x^{1/\epsilon}}{x} dx &\leq A(\epsilon,n) \leq n + \lfloor n^\epsilon \rfloor + \int_1^{n^\epsilon} \frac{n+1-x^{1/\epsilon}}{x} dx
\end{align*} $$
For a general decreasing function, the upper limit of the left integral would be $\lfloor n^\epsilon \rfloor + 1$, but in this case instead it can stop before the integrand becomes negative. On the right, increasing the integral's upper limit only relaxes the inequality.
$$ -\lfloor n^\epsilon \rfloor + n \ln(n^\epsilon) - \epsilon(n - 1) \leq A(\epsilon,n) \leq n + \lfloor n^\epsilon \rfloor + (n+1)\ln(n^\epsilon) - \epsilon(n-1)
 $$
The denominator in the original limit is $A(1,n)$, and
$$ \frac{-\lfloor n^\epsilon \rfloor + \epsilon(n \ln n - n + 1)}{(n+1)\ln n + n+1} \leq \frac{A(\epsilon,n)}{A(1,n)} \leq \frac{n + \lfloor n^\epsilon\rfloor + \epsilon((n+1)\ln n - n + 1)}{n \ln n -2n + 1} $$
Since $n \ln n$ is the dominant term, the limit squeezes to
$$ f(\epsilon) = \lim_{n \to \infty} \frac{A(\epsilon,n)}{A(1,n)} = \epsilon $$
(Note: Dirichlet proved the denominator $\sum_{i=1}^n \Big|\{k \in \mathbb{N} : 1 \leq k \leq i \land  k | i\}\Big|$ (Sloane's A006218) is $n (\ln n + 2 \gamma - 1) + O(\sqrt{n})$, but that much precision wasn't needed here.)
