Evaluate `lim_(xto0) (cos^(-1)((1-x^(2))/(1+x^(2))))/(sin^(-1)x).`

Explanation

To evaluate the limit

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

we will follow these steps:

### Step 1: Rewrite the expression

We start with the limit:

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

### Step 2: Use the trigonometric identity

Using the identity for cosine, we know that:

\[ \frac{1 - \tan^2(\theta)}{1 + \tan^2(\theta)} = \cos(2\theta). \]

We can let \( x = \tan(\theta) \). As \( x \to 0 \), \( \theta \to 0 \) as well. Thus, we rewrite the limit in terms of \( \theta \):

\[ \frac{1 - \tan^2(\theta)}{1 + \tan^2(\theta)} = \cos(2\theta). \]

### Step 3: Substitute \( x = \tan(\theta) \)

Substituting \( x = \tan(\theta) \) into the limit gives us:

\[ \lim_{\theta \to 0} \frac{\cos^{-1}(\cos(2\theta))}{\sin^{-1}(\tan(\theta))}. \]

### Step 4: Simplify the limit

The expression simplifies to:

\[ \lim_{\theta \to 0} \frac{2\theta}{\sin^{-1}(\tan(\theta))}. \]

### Step 5: Use L'Hôpital's Rule

Since both the numerator and denominator approach 0 as \( \theta \to 0 \), we can apply L'Hôpital's Rule:

\[ \lim_{\theta \to 0} \frac{2}{\frac{d}{d\theta}(\sin^{-1}(\tan(\theta)))}. \]

### Step 6: Differentiate the denominator

Using the chain rule, we differentiate \( \sin^{-1}(\tan(\theta)) \):

\[ \frac{d}{d\theta}(\sin^{-1}(\tan(\theta))) = \frac{1}{\sqrt{1 - \tan^2(\theta)}} \cdot \sec^2(\theta). \]

### Step 7: Substitute back into the limit

Now substituting back into our limit gives:

\[ \lim_{\theta \to 0} \frac{2}{\frac{1}{\sqrt{1 - \tan^2(\theta)}} \cdot \sec^2(\theta)}. \]

### Step 8: Evaluate the limit

As \( \theta \to 0 \), \( \tan(\theta) \to 0 \), so:

\[ \sqrt{1 - \tan^2(0)} = 1 \quad \text{and} \quad \sec^2(0) = 1. \]

Thus, the limit simplifies to:

\[ \lim_{\theta \to 0} 2 \cdot 1 = 2. \]

### Final Answer

Therefore, the limit evaluates to:

\[ \boxed{2}. \]