For positive integer n, define $f\left(n\right)=n+\frac{16+5n-3n^2}{4n+3n^2}+\frac{32+n-3n^2}{8n+3n^2}+\frac{48-3n-3n^2}{12n+3n^2}+\mathrm{.......}+\frac{25n-7n^2}{7n^2}$ .
Then the value of $e^{\left(4-\underset{n\to \infty }{\lim }f\left(n\right)\right)}$  is equal to 
 

$$\begin{gathered} f\left(n\right)=n+\sum _{r=1}^n\frac{16r+\left(9-4r\right)n-3n^2}{4rn+3n^2}=n+\sum _{r=1}^n\frac{\left(16r+9n\right)-\left(4n+3n^2\right)}{4rn+3n^2}=\sum _{r=1}^n\frac{\left(16r+9n\right)}{4rn+3n^2} \\ \Rightarrow \underset{n\to \infty }{\lim }f\left(n\right)=\underset{n\to \infty }{\lim }\Sigma \frac{16r+9n}{4rn+3n^2}=\underset{n\to \infty }{\lim }\frac{\left(16\left(\frac{r}{n}\right)+9\right)\frac{1}{n}}{4\left(\frac{r}{n}\right)+3}=\underset{0}{\overset{1}{\int }}\frac{16x+9}{4x+3}dx=\underset{0}{\overset{1}{\int }}4dx-\underset{0}{\overset{1}{\int }}\frac{3dx}{4x+3} \\ =4-\frac{3}{4}{\left(\mathcal{l}n\left|4x+3\right|\right)}_0^1=4-\frac{3}{4}\mathcal{l}n\frac{7}{3} \end{gathered}$$