The value of `lim_(x rarr 0) (1)/(x) [ tan^(-1) ((x+1)/(2x+1)) - (pi)/(4) ]` is

Options

A. 1
B. `-(1)/(2)`
C. 2
D. 0

Answer

To solve the limit

\[ \lim_{x \to 0} \frac{1}{x} \left( \tan^{-1} \left( \frac{x+1}{2x+1} \right) - \frac{\pi}{4} \right), \]

we can follow these steps:

### Step 1: Rewrite the limit expression
We start with the limit expression:

\[ \lim_{x \to 0} \frac{1}{x} \left( \tan^{-1} \left( \frac{x+1}{2x+1} \right) - \frac{\pi}{4} \right). \]

### Step 2: Use the property of the arctangent function
Recall that

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

In our case, we can set \( a = \frac{x+1}{2x+1} \) and \( b = 1 \) (since \(\tan^{-1}(1) = \frac{\pi}{4}\)). Thus, we can rewrite the expression as:

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

### Step 3: Simplify the expression inside the arctangent
Calculating \( \frac{x+1}{2x+1} - 1 \):

\[ \frac{x+1 - (2x + 1)}{2x + 1} = \frac{-x}{2x + 1}. \]

Now, for the denominator:

\[ 1 + \frac{x+1}{2x+1} = \frac{(2x + 1) + (x + 1)}{2x + 1} = \frac{3x + 2}{2x + 1}. \]

Thus, we have:

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

### Step 4: Substitute back into the limit
Now substituting back into the limit, we have:

\[ \lim_{x \to 0} \frac{1}{x} \left( \tan^{-1} \left( \frac{-x}{3x + 2} \right) \right). \]

### Step 5: Apply the limit
As \( x \to 0 \), \( \tan^{-1}(-x/(3x + 2)) \) approaches \( \tan^{-1}(0) = 0 \). We can use the approximation \( \tan^{-1}(u) \approx u \) for small \( u \):

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

Thus, we rewrite the limit:

\[ \lim_{x \to 0} \frac{1}{x} \left( \frac{-x}{3x + 2} \right) = \lim_{x \to 0} \frac{-1}{3x + 2}. \]

### Step 6: Evaluate the limit
Now we can evaluate the limit:

\[ \lim_{x \to 0} \frac{-1}{3x + 2} = \frac{-1}{2}. \]

### Final Answer
Therefore, the value of the limit is:

\[ \boxed{-\frac{1}{2}}. \]