Evaluate $\lim\limits_{n\to \infty}\frac{1}{n+1}+\frac{1}{n+2}+\cdots+\frac{1}{6n}$ Show that $$\lim_{n\to \infty}\left(\frac{1}{n+1}+\frac{1}{n+2}+\cdots+\frac{1}{6n}\right)=\log 6$$ Here I need to use the definition of integral but I  faced problem in range . Please help.

Maybe it is intended that you mention Riemann sums explicitly. 
Rewrite our sum as 
$$\frac{1}{n}\left(\frac{1}{1+\frac{1}{n}} + \frac{1}{1+\frac{2}{n}} + \frac{1}{1+\frac{3}{n}}+\cdots +\frac{1}{1+\frac{5n}{n}} \right).$$
We recognize this as a (right) Riemann sum for the integral
$$\int_0^5 \frac{dx}{1+x},$$
which has value $\log 6$. 
For imagine dividing the interval from $x=0$ to $x=5$ into equal subintervals of width $\frac{1}{n}$. The right ends of these subintervals are at $\frac{1}{n}$, $\frac{2}{n}$, $\frac{3}{n}$, and so on up to $\frac{5n}{n}$. So if $f(x)=\frac{1}{1+x}$, then our summands are precisely the function values at these right endpoints.  The limit of the Riemann sum as $n \to\infty$ is therefore $\log 6$.
