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

Answer

To evaluate the limit

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

we will follow these steps:

### Step 1: Substitute \(x = 0\)

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

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

This gives us an indeterminate form \( \frac{0}{0} \).

**Hint:** If you get \( \frac{0}{0} \), consider using L'Hôpital's rule.

### Step 2: Apply L'Hôpital's Rule

Since we have an indeterminate form, we can apply L'Hôpital's Rule, which states that if \( \lim_{x \to c} \frac{f(x)}{g(x)} = \frac{0}{0} \) or \( \frac{\infty}{\infty} \), then:

\[ \lim_{x \to c} \frac{f(x)}{g(x)} = \lim_{x \to c} \frac{f'(x)}{g'(x)}, \]

provided the limit on the right side exists.

### Step 3: Differentiate the Numerator and Denominator

**Numerator:**
Let \( f(x) = \sqrt{1+x} - \sqrt{1+x^2} \).

Using the chain rule, we differentiate:

\[ f'(x) = \frac{1}{2\sqrt{1+x}} - \frac{1}{2\sqrt{1+x^2}} \cdot (2x) = \frac{1}{2\sqrt{1+x}} - \frac{x}{\sqrt{1+x^2}}. \]

**Denominator:**
Let \( g(x) = \sqrt{1-x^2} - \sqrt{1-x} \).

Differentiating gives:

\[ g'(x) = \frac{-x}{\sqrt{1-x^2}} - \frac{-1}{2\sqrt{1-x}} = \frac{-x}{\sqrt{1-x^2}} + \frac{1}{2\sqrt{1-x}}. \]

### Step 4: Substitute Back into the Limit

Now we substitute \( f'(x) \) and \( g'(x) \) back into the limit:

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

### Step 5: Evaluate the New Limit

Substituting \(x = 0\) into the derivatives:

**Numerator:**

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

**Denominator:**

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

Thus, we have:

\[ \lim_{x \to 0} \frac{f'(x)}{g'(x)} = \frac{\frac{1}{2}}{\frac{1}{2}} = 1. \]

### Final Answer

Therefore, the limit is

\[ \boxed{1}. \]