Evaluate the following limits :
`lim_(x to 0)(sqrt(1-x^(2))-sqrt(1+x^(2)))/(2x^(2))`

Answer

To evaluate the limit

\[ \lim_{x \to 0} \frac{\sqrt{1 - x^2} - \sqrt{1 + x^2}}{2x^2}, \]

we first substitute \(x = 0\) into the expression:

\[ \frac{\sqrt{1 - 0^2} - \sqrt{1 + 0^2}}{2 \cdot 0^2} = \frac{\sqrt{1} - \sqrt{1}}{0} = \frac{0}{0}. \]

Since we have an indeterminate form \( \frac{0}{0} \), we can apply L'Hôpital's Rule, which states that if the limit results in \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \), we can take the derivative of the numerator and the derivative of the denominator.

### Step 1: Differentiate the numerator and denominator

Let \( f(x) = \sqrt{1 - x^2} - \sqrt{1 + x^2} \) and \( g(x) = 2x^2 \).

Now we differentiate \( f(x) \) and \( g(x) \):

1. **For \( f(x) \)**:
- The derivative of \( \sqrt{1 - x^2} \) using the chain rule is:
\[ f'(x) = \frac{1}{2\sqrt{1 - x^2}} \cdot (-2x) = -\frac{x}{\sqrt{1 - x^2}}. \]
- The derivative of \( \sqrt{1 + x^2} \) is:
\[ f'(x) = \frac{1}{2\sqrt{1 + x^2}} \cdot (2x) = \frac{x}{\sqrt{1 + x^2}}. \]
- Therefore, the derivative of the numerator is:
\[ f'(x) = -\frac{x}{\sqrt{1 - x^2}} - \frac{x}{\sqrt{1 + x^2}} = -x \left( \frac{1}{\sqrt{1 - x^2}} + \frac{1}{\sqrt{1 + x^2}} \right). \]

2. **For \( g(x) \)**:
- The derivative of \( g(x) = 2x^2 \) is:
\[ g'(x) = 4x. \]

### Step 2: Apply L'Hôpital's Rule

Now we can rewrite the limit using the derivatives:

\[ \lim_{x \to 0} \frac{f'(x)}{g'(x)} = \lim_{x \to 0} \frac{-x \left( \frac{1}{\sqrt{1 - x^2}} + \frac{1}{\sqrt{1 + x^2}} \right)}{4x}. \]

We can simplify this:

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

### Step 3: Evaluate the limit as \( x \to 0 \)

Substituting \( x = 0 \):

\[ = \frac{-\left( \frac{1}{\sqrt{1 - 0^2}} + \frac{1}{\sqrt{1 + 0^2}} \right)}{4} = \frac{-\left( \frac{1}{1} + \frac{1}{1} \right)}{4} = \frac{-2}{4} = -\frac{1}{2}. \]

Thus, the limit is

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