`lim_(xto0) (1)/(x)cos^(1)((1-x^(2))/(1+x^2))` is equal to

Options

A. `f(x)f(y)`
B. `f(x)+f(y)`
C. `f(x)-f(y)`
D. none of these

Explanation

To solve the limit \( \lim_{x \to 0} \frac{1}{x} \cos^{-1} \left( \frac{1 - x^2}{1 + x^2} \right) \), we can follow these steps:

### Step 1: Simplify the expression inside the cosine inverse
We know that:
\[ \cos^{-1} \left( \frac{1 - x^2}{1 + x^2} \right) = \begin{cases} 2 \tan^{-1}(x) & \text{if } x \geq 0 \\ -2 \tan^{-1}(x) & \text{if } x < 0 \end{cases} \]

### Step 2: Evaluate the left-hand limit as \( x \to 0^- \)
For \( x < 0 \):
\[ \lim_{x \to 0^-} \frac{1}{x} \cos^{-1} \left( \frac{1 - x^2}{1 + x^2} \right) = \lim_{x \to 0^-} \frac{1}{x} (-2 \tan^{-1}(x)) \]
Using the fact that \( \tan^{-1}(x) \approx x \) as \( x \to 0 \):
\[ \lim_{x \to 0^-} \frac{-2 \tan^{-1}(x)}{x} = \lim_{x \to 0^-} \frac{-2x}{x} = -2 \]

### Step 3: Evaluate the right-hand limit as \( x \to 0^+ \)
For \( x \geq 0 \):
\[ \lim_{x \to 0^+} \frac{1}{x} \cos^{-1} \left( \frac{1 - x^2}{1 + x^2} \right) = \lim_{x \to 0^+} \frac{1}{x} (2 \tan^{-1}(x)) \]
Again, using \( \tan^{-1}(x) \approx x \) as \( x \to 0 \):
\[ \lim_{x \to 0^+} \frac{2 \tan^{-1}(x)}{x} = \lim_{x \to 0^+} \frac{2x}{x} = 2 \]

### Step 4: Compare the left-hand limit and right-hand limit
From the calculations:
- Left-hand limit as \( x \to 0^- \) is \( -2 \).
- Right-hand limit as \( x \to 0^+ \) is \( 2 \).

Since the left-hand limit and right-hand limit are not equal, we conclude that:
\[ \lim_{x \to 0} \frac{1}{x} \cos^{-1} \left( \frac{1 - x^2}{1 + x^2} \right) \text{ does not exist.} \]

### Final Answer:
The limit does not exist.