calculus - Value of $lim_{n to infty}frac{sqrt{1}+sqrt{2}+sqrt{3}+cdots+sqrt{n-1}}{nsqrt{n}}$

Evaluate
$$\lim_{n \to \infty}\dfrac{\sqrt{1}+\sqrt{2}+\sqrt{3}+\cdots+\sqrt{n-1}}{n\sqrt{n}}.$$

I am trying to use the Sandwich principle here..

$\lim_{n \to \infty}\dfrac{\sqrt{1}+\sqrt{2}+\sqrt{3}+\cdots+\sqrt{n-1}}{n\sqrt{n}}=\lim_{n \to \infty}\dfrac{\sqrt{\dfrac{1}{n}}+\sqrt{\dfrac{2}{n}}+\sqrt{\dfrac{3}{n}}+\cdots+\sqrt{\dfrac{n-1}{n}}}{n}\ge \lim_{n \to \infty}\bigg(\sqrt{\dfrac{1}{n}}.\sqrt{\dfrac{2}{n}}.\sqrt{\dfrac{3}{n}}\cdots\sqrt{\dfrac{n-1}{n}}\bigg )^\dfrac{1}{n-1}=\lim_{n \to \infty}\bigg(\dfrac{(n-1)!}{n^n}\bigg )^\dfrac{1}{2(n-1)}$

But after this I am a little in doubt.
This link may provide some light Evaluation of the limit $\lim\limits_{n \to \infty } \frac1{\sqrt n}\left(1 + \frac1{\sqrt 2 }+\frac1{\sqrt 3 }+\cdots+\frac1{\sqrt n } \right)$ but I do not understand how it would help my problem..

In continuation of @Rebello's answer here, I would like to provide an answer for the problem given in the link

$\lim\limits_{n \to \infty } \frac1{\sqrt n}\left(1 + \frac1{\sqrt 2 }+\frac1{\sqrt 3 }+\cdots+\frac1{\sqrt n } \right)=\lim\limits_{n \to \infty }\sum_{k=1}^{n}{\dfrac{1}{\sqrt{kn}}}=\int_{0}^{1}\dfrac{1}{\sqrt{x}}dx+\lim\limits_{n \to \infty }\dfrac{1}{n}=2$

Answer

Hint :

$$\lim_{n \to \infty}\dfrac{\sqrt{1}+\sqrt{2}+\sqrt{3}+...+\sqrt{n-1}}{n\sqrt{n}} = \lim_{n \to \infty} \frac{1}{n} \sum_{k=0}^{n-1} \sqrt{\frac{k}{n}} = \int_0^1\sqrt{x}\mathrm{d}x$$