The correct option is (B): \(4-\frac{3}{4} \log _e\left(\frac{7}{3}\right)\)
We are given the function:
\(f(n) = n + \frac{16 + 5n - 3n^2}{4n + 3n^2} + \cdots + \frac{25n - 7n^2}{7n^2}\)
\(= \left(\frac{16 + 5n - 3n^2}{4n + 3n^2} + 1\right) + \left(\frac{32 + n - 3n^2}{8n + 3n^2} + 1\right) + \cdots + \left(\frac{25n - 7n^2}{7n^2} + 1\right)\)
We need to find the limit:
\(\lim_{n \to \infty}f(n)\)
Now let us simplify
\(f(n) = \frac{9n + 16}{4n + 3n^2} + \frac{9n + 32}{8n + 3n^2} + \cdots + \frac{25n}{7n^2}\)
\(= \sum_{r=1}^{n} \frac{9n + 16r}{4rn + 3n^2} = \frac{1}{n} \sum_{r=1}^{n} \frac{9 + 16\left(\frac{r}{n}\right)}{4 \left(\frac{r}{n}\right) + 3}\)
\(\lim_{n \to \infty} f(n) = \int_{0}^{1} \frac{9 + 16x}{4x + 3} \, dx\)
\(= \int_{0}^{1} \left(\frac{16x + 12}{4x + 3} - 3\right) \, dx\)
\(= \left[4x - \frac{3}{4} \ln |4x + 3|\right]_{0}^{1}\)
\(=4-\frac{3}{4} \log _e\left(\frac{7}{3}\right)\)
So, the correct option is (B): \(4-\frac{3}{4} \log _e\left(\frac{7}{3}\right)\)
A function's limit is a number that a function reaches when its independent variable comes to a certain value. The value (say a) to which the function f(x) approaches casually as the independent variable x approaches casually a given value "A" denoted as f(x) = A.
If limx→a- f(x) is the expected value of f when x = a, given the values of ‘f’ near x to the left of ‘a’. This value is also called the left-hand limit of ‘f’ at a.
If limx→a+ f(x) is the expected value of f when x = a, given the values of ‘f’ near x to the right of ‘a’. This value is also called the right-hand limit of f(x) at a.
If the right-hand and left-hand limits concur, then it is referred to as a common value as the limit of f(x) at x = a and denote it by lim x→a f(x).