$\sqrt{x^2.x}$ always has decimal portion starting ".000...49"

Expanding $\sqrt{n^2+h}$ in Taylor series around $h=0$ gives $$ \sqrt{n^2+h}=n+\frac{h}{2n}-\frac{h^2}{8n^3}+O(h^3) $$ When $h=\dfrac{n}{10}$, we get $$ \sqrt{n^2+\frac{n}{10}} \approx n+\frac{1}{20}-\frac{1}{800n} $$ So, the fractional part is a bit less than $1/20=0.05$, and the bit is less than $1/800<0.001$ for $n\ge 2$. Therefore, the fractional part is $0.049\cdots$.

The same argument holds for larger $n$ with the proper power of $10$.