Probability of selecting positive integer with odd number of digits (Grinstead and Snell Ch 1.2 Q 28 (d)) If $A$ is the set of all positive integers with an odd number of digits, let $A\,(N)$ be the number of elements of $A$ which are less than or equal to $N$. Show that $P\,(A) = \lim_{N\to\infty} A\,(N)/N$ does not exist.

My rusty knowledge of limits says $P\,(A) = 0$, since the denominator dominates. Any help?
Hint:

Compute $A(10^{2n})/10^{2n}$ and $A(10^{2n+1})/10^{2n+1}$.
If you look at the beginning of the even set with $N_k=10^{2k+1}$ where k is an Integer then the last $10^{2k+1}-10^{2k}$ have been elements with an odd number of digits. Therefore we get $P(N_k) = A(N_k)/N_k \geq \frac{10^{2k+1}-10^{2k}}{10^{2k+1}}  $ which goes to 0.9 for k to infinity. But we can also look at the $M_k=10^{2k}$. For this we can show that $P(M_k)$ goes to 0.1 . Therefore the limit can‘t converge since we found 2 different  accumulation points.