`lim_(n to oo)sum_(r=1)^(n)cot^(-1)(r^(2)+3//4)` is (a) 0 (b) `tan^(-1)1` (c) `tan^(-1)2` (d) None of these

Options

A. 0
B. `tan^(-1)1`
C. `tan^(-1)2`
D. None of these

Answer

To solve the limit \( \lim_{n \to \infty} \sum_{r=1}^{n} \cot^{-1}\left(r^2 + \frac{3}{4}\right) \), we can follow these steps:

### Step 1: Rewrite the cotangent inverse
We start by rewriting \( \cot^{-1}(x) \) in terms of \( \tan^{-1}(x) \):
\[ \cot^{-1}(x) = \frac{\pi}{2} - \tan^{-1}(x) \]
Thus, we can express our limit as:
\[ \lim_{n \to \infty} \sum_{r=1}^{n} \left( \frac{\pi}{2} - \tan^{-1}\left(r^2 + \frac{3}{4}\right) \right) \]

### Step 2: Separate the sum
This can be separated into two parts:
\[ \lim_{n \to \infty} \left( \sum_{r=1}^{n} \frac{\pi}{2} - \sum_{r=1}^{n} \tan^{-1}\left(r^2 + \frac{3}{4}\right) \right) \]

### Step 3: Evaluate the first sum
The first sum is straightforward:
\[ \sum_{r=1}^{n} \frac{\pi}{2} = n \cdot \frac{\pi}{2} \]
As \( n \to \infty \), this term approaches \( \infty \).

### Step 4: Analyze the second sum
Now, we need to analyze the second sum:
\[ \sum_{r=1}^{n} \tan^{-1}\left(r^2 + \frac{3}{4}\right) \]
As \( r \) becomes large, \( \tan^{-1}(r^2 + \frac{3}{4}) \) approaches \( \frac{\pi}{2} \). Therefore, for large \( n \):
\[ \tan^{-1}\left(r^2 + \frac{3}{4}\right) \approx \frac{\pi}{2} \]
Thus, the sum behaves like:
\[ \sum_{r=1}^{n} \tan^{-1}\left(r^2 + \frac{3}{4}\right) \approx n \cdot \frac{\pi}{2} \]

### Step 5: Combine the results
Now, we can combine the results of both sums:
\[ \lim_{n \to \infty} \left( n \cdot \frac{\pi}{2} - n \cdot \frac{\pi}{2} \right) = \lim_{n \to \infty} 0 = 0 \]

### Final Result
Thus, the limit evaluates to:
\[ \boxed{0} \]