First, let's write the expression as a sum:
$s_n=\sum\limits_{k = 1}^n {\frac{1}{{\sqrt {n + k} }}\frac{1}{{\sqrt n }}} $
It is first stated that
$\sum\limits_{k = 1}^n {f\left( {\frac{k}{n}} \right)\frac{1}{n} \to \int\limits_0^1 {f\left( x \right)dx} } $
This means that the sum constructed on the left will tend to the value of the definite integral of $f$ over $[0,1]$. This is a result from Darboux/Riemann integration you might find in most textbooks. Assuming this result, we seek to use it to evaluate some sums. First, we need to write
$s_n=\sum\limits_{k = 1}^n {\frac{1}{{\sqrt {n + k} }}\frac{1}{{\sqrt n }}} $
as
$\sum\limits_{k = 1}^n f\left( {\frac{k}{n}} \right) \frac{1}{n}$
for some $f$. To find $f$, we must isolate the $1/n$ term, and see what is left. In this case:
$\eqalign{ & {s_n} = \sum\limits_{k = 1}^n {\frac{1}{{\sqrt {n + k} }}\frac{1}{{\sqrt n }}} = \sum\limits_{k = 1}^n {\frac{1}{n}\frac{1}{{\sqrt {n + k} }}\frac{n}{{\sqrt n }}} = \cr & {s_n} = \sum\limits_{k = 1}^n {\frac{1}{n}\frac{{\sqrt n }}{{\sqrt {n + k} }}} = \sum\limits_{k = 1}^n {\frac{1}{n}\sqrt {\frac{n}{{n + k}}} } = \sum\limits_{k = 1}^n {\frac{1}{n}\frac{1}{{\sqrt {\frac{{n + k}}{n}} }}} = \sum\limits_{k = 1}^n {\frac{1}{n}\frac{1}{{\sqrt {1 + \frac{k}{n}} }}} \cr} $
Can you take it from there?
Do not hover over the grey areas unless you want a solution. Try to think about it first.
So we can see that $f(x)=\frac{1}{\sqrt{1+x}}$. This means that $\mathop {\lim }\limits_{n \to \infty } {s_n} = \mathop {\lim }\limits_{n \to \infty } \sum\limits_{k = 1}^n {\frac{1}{{\sqrt n }}\frac{1}{{\sqrt {n + k} }}} = \mathop {\lim }\limits_{n \to \infty } \sum\limits_{k = 1}^n {\frac{1}{n}\frac{1}{{\sqrt {1 + \frac{k}{n}} }}} = \int\limits_0^1 {\frac{{dx}}{{\sqrt {1 + x} }}} = \frac{1}{2}\left( {1 - \sqrt 2 } \right)$