Following the first term [tex]n[/tex], the [tex]k[/tex]-th term [tex]\left(k\in\{1,2,3,\ldots,n\}\right)[/tex] in [tex]f(n)[/tex] is given by
[tex]\dfrac{16k + (9-4k)n - 3n^2}{4kn + 3n^2}[/tex]
That is,
[tex]k=1 \implies \dfrac{16 + (9-4)n - 3n^2}{4n + 3n^2} = \dfrac{16 + 5n - 3n^2}{4n + 3n^2}[/tex]
[tex]k=2 \implies \dfrac{16\cdot2 + (9-8)n - 3n^2}{4\cdot2n + 3n^2} = \dfrac{32 + n - 3n^2}{8n + 3n^2}[/tex]
and so on, up to
[tex]k=n \implies \dfrac{16n + (9-4n)n - 3n^2}{4n^2 + 3n^2} = \dfrac{25n - 7n^2}{7n^2}[/tex]
We then have
[tex]\displaystyle f(n) = n + \sum_{k=1}^n \frac{16k + (9-4k)n - 3n^2}{4kn + 3n^2} \\\\ ~~~~~~~ = n + \frac1n \sum_{k=1}^n \frac{16\frac kn + 9}{4\frac kn + 3} - \frac1n \sum_{k=1}^n \frac{4kn + 3n^2}{4k + 3n} \\\\ ~~~~~~~ = n + \frac1n \sum_{k=1}^n \frac{16\frac kn + 9}{4\frac kn + 3}- \frac1n \sum_{k=1}^n n \\\\ ~~~~~~~ = \frac1n \sum_{k=1}^n \frac{16\frac kn + 9}{4\frac kn + 3}[/tex]
Now as [tex]n\to\infty[/tex], the right side converges to a definite integral.
[tex]\displaystyle \lim_{n\to\infty} f(n) = \lim_{n\to\infty} \frac1n \sum_{k=1}^n \frac{16\frac kn + 9}{4\frac kn + 3} \\\\ ~~~~~~~~~~~~~ \,= \int_0^1 \frac{16x + 9}{4x + 3} \, dx[/tex]
Wrap up by substituting [tex]x\mapsto\frac{x-3}4[/tex].
[tex]\displaystyle \lim_{n\to\infty} f(n) = \frac14 \int_0^1 \frac{16x + 9}{4x + 3} \, dx \\\\ ~~~~~~~~~~~~~\, = \frac14 \int_3^7 \frac{4x - 3}x \, dx \\\\ ~~~~~~~~~~~~~\, = \frac14 \int_3^7 \left(4 - \frac3x\right) \, dx \\\\ ~~~~~~~~~~~~~\, = \left. \frac14 (4x - 3\ln|x|) \right\vert_3^7\\\\ ~~~~~~~~~~~~~\, = \boxed{4 + \frac34 \ln\left(\frac37\right)}[/tex]
