`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 inverse cosine
We know that:
\[ \cos^{-1}\left(\frac{1 - x^2}{1 + x^2}\right) \]
can be rewritten using the identity:
\[ \cos^{-1}(y) = 2 \tan^{-1}(\sqrt{\frac{1-y}{1+y}}) \]
However, for small values of \(x\), we can use the known result:
\[ \cos^{-1}\left(\frac{1 - x^2}{1 + x^2}\right) = 2 \tan^{-1}(x) \]
This holds true for \(x \geq 0\) and \(x < 0\) with a sign adjustment.

### Step 2: Substitute the identity into the limit
Now, substituting this into our limit gives:
\[ \lim_{x \to 0} \frac{1}{x} \cdot 2 \tan^{-1}(x) \]

### Step 3: Split the limit into two cases
We will evaluate the limit from the right-hand side (as \(x\) approaches 0 from positive values) and the left-hand side (as \(x\) approaches 0 from negative values).

#### Right-Hand Limit (as \(x \to 0^+\)):
\[ \lim_{x \to 0^+} \frac{2 \tan^{-1}(x)}{x} \]
Using the known limit:
\[ \lim_{x \to 0} \frac{\tan^{-1}(x)}{x} = 1 \]
Thus:
\[ \lim_{x \to 0^+} \frac{2 \tan^{-1}(x)}{x} = 2 \cdot 1 = 2 \]

#### Left-Hand Limit (as \(x \to 0^-\)):
For \(x < 0\), we have:
\[ \lim_{x \to 0^-} \frac{1}{x} \cdot (-2 \tan^{-1}(x)) = \lim_{x \to 0^-} \frac{-2 \tan^{-1}(x)}{x} \]
Again, using the same limit:
\[ \lim_{x \to 0} \frac{\tan^{-1}(x)}{x} = 1 \]
Thus:
\[ \lim_{x \to 0^-} \frac{-2 \tan^{-1}(x)}{x} = -2 \cdot 1 = -2 \]

### Step 4: Conclusion
Since the right-hand limit (2) and the left-hand limit (-2) are not equal, we conclude that:
\[ \lim_{x \to 0} \frac{1}{x} \cos^{-1}\left(\frac{1 - x^2}{1 + x^2}\right) \text{ does not exist.} \]