Find the value of `lim_(n to oo) (tan(sum_(r=1)^(n) tan^(-1)((4)/(4r^(2)+3))))`.

Answer

To find the value of

\[ \lim_{n \to \infty} \tan\left(\sum_{r=1}^{n} \tan^{-1}\left(\frac{4}{4r^2+3}\right)\right), \]

we will follow these steps:

### Step 1: Rewrite the summation
We start by rewriting the term inside the summation:

\[ \tan^{-1}\left(\frac{4}{4r^2 + 3}\right) = \tan^{-1}\left(\frac{4}{4(r^2 + \frac{3}{4})}\right) = \tan^{-1}\left(\frac{1}{r^2 + \frac{3}{4}}\right). \]

### Step 2: Use the tangent subtraction formula
Using the identity for the tangent of a difference, we can express the summation as:

\[ \sum_{r=1}^{n} \tan^{-1}(x) - \tan^{-1}(y) = \tan^{-1}\left(\frac{x - y}{1 + xy}\right). \]

We can set \( x = r + \frac{1}{2} \) and \( y = r - \frac{1}{2} \) to apply this formula.

### Step 3: Express the summation
Thus, we can express the summation as:

\[ \sum_{r=1}^{n} \left(\tan^{-1}\left(r + \frac{1}{2}\right) - \tan^{-1}\left(r - \frac{1}{2}\right)\right). \]

### Step 4: Telescoping series
This forms a telescoping series, where most terms will cancel out. The remaining terms will be:

\[ \tan^{-1}\left(n + \frac{1}{2}\right) - \tan^{-1}\left(\frac{1}{2}\right). \]

### Step 5: Take the tangent
Now we need to find:

\[ \tan\left(\tan^{-1}\left(n + \frac{1}{2}\right) - \tan^{-1}\left(\frac{1}{2}\right)\right). \]

Using the tangent subtraction formula again, we have:

\[ \tan(a - b) = \frac{\tan a - \tan b}{1 + \tan a \tan b}. \]

### Step 6: Substitute values
Substituting \( a = \tan^{-1}(n + \frac{1}{2}) \) and \( b = \tan^{-1}(\frac{1}{2}) \):

\[ \tan\left(\tan^{-1}\left(n + \frac{1}{2}\right)\right) = n + \frac{1}{2}, \]
\[ \tan\left(\tan^{-1}\left(\frac{1}{2}\right)\right) = \frac{1}{2}. \]

So we have:

\[ \tan\left(\tan^{-1}\left(n + \frac{1}{2}\right) - \tan^{-1}\left(\frac{1}{2}\right)\right) = \frac{(n + \frac{1}{2}) - \frac{1}{2}}{1 + (n + \frac{1}{2}) \cdot \frac{1}{2}} = \frac{n}{1 + \frac{n}{2} + \frac{1}{4}}. \]

### Step 7: Simplify the expression
This simplifies to:

\[ \frac{n}{1 + \frac{n}{2} + \frac{1}{4}} = \frac{n}{\frac{4 + 2n + 1}{4}} = \frac{4n}{2n + 5}. \]

### Step 8: Take the limit
Now we take the limit as \( n \to \infty \):

\[ \lim_{n \to \infty} \frac{4n}{2n + 5} = \lim_{n \to \infty} \frac{4}{2 + \frac{5}{n}} = \frac{4}{2} = 2. \]

### Final Answer
Thus, the value of the limit is

\[ \boxed{2}. \]