The value of `lim_(x to oo)(d)/(dx)int_(sqrt(3))^(sqrt(x))(r^(3))/((r+1)(r-1))dr` is :

Options

A. 0
B. 1
C. `(1)/(2)`
D. None of these

Answer

To solve the given limit problem, we will follow these steps:

### Step 1: Set Up the Expression
We start with the expression we need to evaluate:
\[ \lim_{x \to \infty} \frac{d}{dx} \int_{\sqrt{3}}^{\sqrt{x}} \frac{r^3}{(r+1)(r-1)} \, dr \]

### Step 2: Apply the Fundamental Theorem of Calculus
According to the Fundamental Theorem of Calculus, we can differentiate the integral:
\[ \frac{d}{dx} \int_{a}^{g(x)} f(r) \, dr = f(g(x)) \cdot g'(x) \]
where \( g(x) = \sqrt{x} \) and \( a = \sqrt{3} \). Thus, we have:
\[ \frac{d}{dx} \int_{\sqrt{3}}^{\sqrt{x}} \frac{r^3}{(r+1)(r-1)} \, dr = \frac{\left(\sqrt{x}\right)^3}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)} \cdot \frac{d}{dx}(\sqrt{x}) \]

### Step 3: Compute \( g'(x) \)
The derivative of \( g(x) = \sqrt{x} \) is:
\[ g'(x) = \frac{1}{2\sqrt{x}} \]

### Step 4: Substitute \( g(x) \) and \( g'(x) \) into the Expression
Now substituting back, we get:
\[ \frac{d}{dx} \int_{\sqrt{3}}^{\sqrt{x}} \frac{r^3}{(r+1)(r-1)} \, dr = \frac{x^{3/2}}{(\sqrt{x}+1)(\sqrt{x}-1)} \cdot \frac{1}{2\sqrt{x}} \]

### Step 5: Simplify the Expression
Simplifying this expression:
\[ = \frac{x^{3/2}}{(\sqrt{x}+1)(\sqrt{x}-1)} \cdot \frac{1}{2\sqrt{x}} = \frac{x^{3/2}}{2\sqrt{x}((\sqrt{x}+1)(\sqrt{x}-1))} = \frac{x^{3/2}}{2\sqrt{x}(x-1)} = \frac{x}{2(x-1)} \]

### Step 6: Evaluate the Limit
Now we need to evaluate the limit:
\[ \lim_{x \to \infty} \frac{x}{2(x-1)} \]
This simplifies to:
\[ = \lim_{x \to \infty} \frac{x}{2x - 2} = \lim_{x \to \infty} \frac{1}{2 - \frac{2}{x}} = \frac{1}{2} \]

### Final Answer
Thus, the value of the limit is:
\[ \frac{1}{2} \]