$\underset{n\to \infty }{Lt} \left[\frac{{\displaystyle 1}}{{\displaystyle na}}+\frac{{\displaystyle 1}}{{\displaystyle na+1}}+\frac{{\displaystyle 1}}{{\displaystyle na+2}}+\mathrm{.........}+\frac{{\displaystyle 1}}{{\displaystyle nb}}\right]=$

$\underset{n\to \infty }{lt}\left[\frac{1}{na}+\frac{1}{na+1}+\frac{1}{na+2}+\mathrm{.....}+\frac{1}{nb}\right]$

$=\underset{n\to \infty }{lt} \left[\frac{{\displaystyle 1}}{{\displaystyle na}}+\frac{{\displaystyle 1}}{{\displaystyle na+1}}+\frac{{\displaystyle 1}}{{\displaystyle na+2}}+\mathrm{.....}+\frac{{\displaystyle 1}}{{\displaystyle na+n(b-a)}}\right]$

$=\underset{n\to \infty }{lt}\sum _{r=0}^{(b-a)n}\frac{1}{na+r}$

$=\underset{n\to \infty }{lt}\frac{1}{n}\sum _{r=0}^{(b-a)n}\frac{1}{a+\left(\frac{r}{n}\right)}$

$=\underset{0}{\overset{(b-a)}{\int }}\frac{1}{a+x}$

$={\left(\log (a+x)\right)}_0^{(b-a)}=\log (a+b-a)-\log a=\log \left(\frac{{\displaystyle b}}{{\displaystyle a}}\right)$

$Formula \underset{n\to \infty }{lt}\frac{1}{n}\sum _{r=0}^{nk}f\left(\frac{r}{n}\right)=\underset{0}{\overset{k}{\int }}f\left(x\right)dx$