find the value of `lim_(x to oo) 48x (pi/4 - tan^(-1) ((x+1)/(x+2)))`

Answer

To find the limit

\[ \lim_{x \to \infty} 48x \left( \frac{\pi}{4} - \tan^{-1} \left( \frac{x+1}{x+2} \right) \right), \]

we can follow these steps:

### Step 1: Rewrite \(\frac{\pi}{4}\) as \(\tan^{-1}(1)\)

We know that \(\tan^{-1}(1) = \frac{\pi}{4}\). Thus, we can rewrite the limit as:

\[ \lim_{x \to \infty} 48x \left( \tan^{-1}(1) - \tan^{-1} \left( \frac{x+1}{x+2} \right) \right). \]

### Step 2: Use the formula for the difference of arctangents

Using the formula for the difference of arctangents:

\[ \tan^{-1}(a) - \tan^{-1}(b) = \tan^{-1} \left( \frac{a - b}{1 + ab} \right), \]

we set \(a = 1\) and \(b = \frac{x+1}{x+2}\). This gives us:

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

### Step 3: Simplify the expression inside the arctangent

Now we simplify the expression:

1. The numerator becomes:
\[ 1 - \frac{x+1}{x+2} = \frac{(x+2) - (x+1)}{x+2} = \frac{1}{x+2}. \]

2. The denominator becomes:
\[ 1 + \frac{x+1}{x+2} = \frac{(x+2) + (x+1)}{x+2} = \frac{2x + 3}{x + 2}. \]

Thus, we have:

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

### Step 4: Substitute back into the limit

Now substituting back into the limit, we have:

\[ \lim_{x \to \infty} 48x \tan^{-1} \left( \frac{1}{2x + 3} \right). \]

### Step 5: Use the limit property of \(\tan^{-1}\)

As \(x \to \infty\), \(\frac{1}{2x + 3} \to 0\). We can use the limit property:

\[ \lim_{u \to 0} \frac{\tan^{-1}(u)}{u} = 1. \]

Thus, we can rewrite:

\[ \tan^{-1} \left( \frac{1}{2x + 3} \right) \sim \frac{1}{2x + 3} \text{ as } x \to \infty. \]

### Step 6: Substitute and simplify

Now we substitute this back into our limit:

\[ \lim_{x \to \infty} 48x \cdot \frac{1}{2x + 3} = \lim_{x \to \infty} \frac{48x}{2x + 3}. \]

### Step 7: Divide numerator and denominator by \(x\)

Dividing both the numerator and denominator by \(x\):

\[ \lim_{x \to \infty} \frac{48}{2 + \frac{3}{x}}. \]

### Step 8: Evaluate the limit

As \(x \to \infty\), \(\frac{3}{x} \to 0\):

\[ \lim_{x \to \infty} \frac{48}{2 + 0} = \frac{48}{2} = 24. \]

### Final Answer

Thus, the value of the limit is:

\[ \boxed{24}. \]