If $D(m)$ is the deficiency of the deficient number $m$, then what is $\lim_{m \rightarrow \infty}{\frac{D(m)}{m}}$? Let $\sigma=\sigma_1$ denote the classical sum-of-divisors function. Call the function $D(x)=2x-\sigma(x)$ as the _deficiency_ of $x$. Here is my question: > If $D(m)$ is the deficiency of the **deficient** number $m$, then what is $$\lim_{m \rightarrow \infty}{\frac{D(m)}{m}}?$$

**Hint** : Both primes and powers of two are deficient, and the limit of these subseqeunces are $1$ and $0$, respectively.