Limit of the decimal part of $\sqrt{4n^2+38n}$

Question

Let $x_n=4n^2+38n$. Find the limit of the decimal part of $\sqrt{x_n}$ as $n$ goes to infinity.

I have tried completing the square for $x_n$, which yields $\sqrt{x_n}<2n+\frac{19}{2}$, but that's all I got. The decimal part really makes the problem harder.

Answer 1

We have $$\lim_{n\to\infty}\sqrt{4n^2+38n+\frac{361}4}-\sqrt{4n^2+38n}=\lim_{n\to\infty}\frac{\frac{361}4}{\sqrt{4n^2+38n+\frac{361}4}+\sqrt{4n^2+38n}}=0.$$ We also have $$\left\{\sqrt{4n^2+38n+\frac{361}4}\right\}=\left\{2n+\frac{19}2\right\}=\frac12,$$ where $\{\cdot\}$ denotes the fractional part. Can you take it from here?

Answer 2

Using binomial expansion for square root,

$\sqrt{4n^2+38n}=2n\sqrt{1+\dfrac{19}{2n}}=2n\left(1+\dfrac{19}{4n}+O\left(\dfrac1{n^2}\right)\right)\to2n+\dfrac{19}2=2n+9+\dfrac12$,

so the decimal part goes to $\dfrac12$.