If $\text{⨍ : R→R}$ is given by $\text{⨍ (x)=x+1,}$ then the value 

$\underset{\mathrm{n}\to \infty }{\lim }\frac{1}{\mathrm{n}}\left[\text{⨍ (0)+⨍ }\left(\frac{5}{\mathrm{n}}\right)\text{+⨍ }\left(\frac{10}{\mathrm{n}}\right)\text{+……+⨍ }\left(\frac{5\left(\mathrm{n}-1\right)}{\mathrm{n}}\right)\right],$ is

$$\begin{gathered} \text{⨍ (x)=x+1} \\ \underset{\mathrm{n}\to \infty }{\mathrm{Lt}}\frac{1}{\mathrm{n}}\left[\text{f0)+f}\left(\frac{5}{\mathrm{n}}\right)\text{+f}\left(\frac{10}{\mathrm{n}}\right)\text{+……+f}\left(\frac{5(\mathrm{n}-1)}{\mathrm{n}}\right)\right] \\ =\underset{\mathrm{n}\to \infty }{\mathrm{Lt}}\frac{1}{\mathrm{n}}\left[1+\frac{5}{\mathrm{n}}+1+\frac{10}{\mathrm{n}}+1+\dots \dots +\frac{5(\mathrm{n}-1)}{\mathrm{n}}+1\right] \\ =\underset{\mathrm{n}\to \infty }{\mathrm{Lt}}\frac{1}{\mathrm{n}}\left[\frac{{\mathrm{n}}^2+5+10+\dots \dots +5(\mathrm{n}-1)}{\mathrm{n}}\right] \\ =\underset{\mathrm{n}\to \infty }{\mathrm{Lt}}\frac{1}{\mathrm{n}}\left[\frac{{\mathrm{n}}^2+5(1+2+\dots \dots +(\mathrm{n}-1\left)\right)}{\mathrm{n}}\right] \\ =\underset{\mathrm{n}\to \infty }{\mathrm{Lt}}\frac{{\mathrm{n}}^2+{\displaystyle \frac{5\mathrm{n}(\mathrm{n}-1)}{2}}}{{\mathrm{n}}^2} \\ =1+\frac{5}{2}=\frac{7}{2} \end{gathered}$$