$\mathop {\lim }\limits_{n \to \infty } {1 \over {\sqrt n }} \left({1 \over {\sqrt 1 }} + {1 \over {\sqrt 2 }} +\cdots+{1 \over {\sqrt n }}\right)$ $$\mathop {\lim }\limits_{n \to \infty } {1 \over {\sqrt n }} \left({1 \over {\sqrt 1 }} + {1 \over {\sqrt 2 }} + {1 \over {\sqrt 3 }}+\cdots+{1 \over {\sqrt n }}\right)$$
( Without use of integrals ).
I was able to squeeze it from the bottom to ${\lim }=1$, but that's not good enough.
I'd be glad for help.

$$ \sqrt{n+1} - \sqrt{n} = \frac{1}{\sqrt{n+1} + \sqrt{n}} < \frac{1}{{\sqrt{n} + \sqrt{n}}} < \frac{1}{2\sqrt{n}}$$
add inequalities from 1 to n,
$$\sqrt{n+1} - 1 > \frac{1}{2}\left( \frac{1}{\sqrt{1}} + \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{3}} + \cdots+ \frac{1}{\sqrt{n}}\right)$$
also for lower bound note that, 
$$\left( \frac{1}{\sqrt{1}} + \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{3}} + \cdots+ \frac{1}{\sqrt{n}}\right) > \left( \frac{1}{\sqrt{n}} + \frac{1}{\sqrt{n}} + \frac{1}{\sqrt{n}} + \cdots+ \frac{1}{\sqrt{n}}\right)= \frac{n}{\sqrt{n}}$$
so we have $$2(\sqrt{n+1} - 1) > \left( \frac{1}{\sqrt{1}} + \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{3}} + \cdots+ \frac{1}{\sqrt{n}}\right) > \sqrt{n}$$
Now sandwich theorem tells us that limit is 2
EDIT:
$$\frac{1}{\sqrt{n+1} + \sqrt{n+1}} < \frac{1}{\sqrt{n}+\sqrt{n+1}}  < \frac{1}{\sqrt{n}+\sqrt{n}}$$
$$\frac{1}{\sqrt{n+1} + \sqrt{n+1}}< \sqrt{n+1} - \sqrt{n} < \frac{1}{\sqrt{n}+\sqrt{n}} $$
add from 1 to n and let $$S = \left( \frac{1}{\sqrt{1}} + \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{3}} + \cdots+ \frac{1}{\sqrt{n}}\right)$$,
then, 
$$\frac{1}{2}\left( S + \frac{1}{\sqrt{n+1}} -1\right) < \sqrt{n+1} -1 < \frac{1}{2}S$$
$$2(\sqrt{n+1} -1) <S<2(\sqrt{n+1} -1) +1 -\frac{1}{\sqrt{n+1}}$$
