The value of `lim_(xto oo)(x+2)tan^(-1)(x+2)-(xtan^(-1)x)` is

Options

A. `(pi)/2`
B. Doesn't exist
C. `(pi)/4`
D. None of these

Answer

To solve the limit problem \( \lim_{x \to \infty} \left( (x + 2) \tan^{-1}(x + 2) - x \tan^{-1}(x) \right) \), we can follow these steps:

### Step 1: Rewrite the limit expression
We start with the expression:
\[ \lim_{x \to \infty} \left( (x + 2) \tan^{-1}(x + 2) - x \tan^{-1}(x) \right) \]
We can expand the first term:
\[ = \lim_{x \to \infty} \left( x \tan^{-1}(x + 2) + 2 \tan^{-1}(x + 2) - x \tan^{-1}(x) \right) \]
This simplifies to:
\[ = \lim_{x \to \infty} \left( x \left( \tan^{-1}(x + 2) - \tan^{-1}(x) \right) + 2 \tan^{-1}(x + 2) \right) \]

### Step 2: Analyze the limit of the arctangent difference
Next, we focus on the term \( \tan^{-1}(x + 2) - \tan^{-1}(x) \):
Using the property of the arctangent function:
\[ \tan^{-1}(a) - \tan^{-1}(b) = \tan^{-1}\left( \frac{a - b}{1 + ab} \right) \]
We can apply this:
\[ \tan^{-1}(x + 2) - \tan^{-1}(x) = \tan^{-1}\left( \frac{(x + 2) - x}{1 + (x + 2)x} \right) = \tan^{-1}\left( \frac{2}{1 + x^2 + 2x} \right) \]

### Step 3: Substitute back into the limit
Substituting this back into our limit gives:
\[ \lim_{x \to \infty} \left( x \tan^{-1}\left( \frac{2}{1 + x^2 + 2x} \right) + 2 \tan^{-1}(x + 2) \right) \]

### Step 4: Evaluate the limit of the arctangent term
As \( x \to \infty \), \( \frac{2}{1 + x^2 + 2x} \to 0 \), thus:
\[ \tan^{-1}\left( \frac{2}{1 + x^2 + 2x} \right) \to 0 \]
Therefore, the first term becomes:
\[ x \cdot 0 \to 0 \]

### Step 5: Evaluate the limit of the second term
Now we evaluate the second term:
\[ \lim_{x \to \infty} 2 \tan^{-1}(x + 2) = 2 \cdot \frac{\pi}{2} = \pi \]

### Step 6: Combine the results
Putting it all together, we have:
\[ \lim_{x \to \infty} \left( (x + 2) \tan^{-1}(x + 2) - x \tan^{-1}(x) \right) = 0 + \pi = \pi \]

### Final Answer
Thus, the value of the limit is:
\[ \pi \]