Lim_x->0 (sqrt(1+x2) - sqrt(1-x)) / x2

Numerator: $\sqrt{1+0^2} - \sqrt{1-0} = \sqrt{1} - \sqrt{1} = 1-1 = 0$.
Denominator: $0^2 = 0$.
Since the result is $\frac{0}{0}$, it is an indeterminate form, and further steps are needed.

The conjugate of $\sqrt{1+x^2} - \sqrt{1-x}$ is $\sqrt{1+x^2} + \sqrt{1-x}$.

Numerator: $0+1 = 1$.
Denominator: $0 \times (\sqrt{1+0^2} + \sqrt{1-0}) = 0 \times (\sqrt{1} + \sqrt{1}) = 0 \times (1+1) = 0 \times 2 = 0$.
The expression is of the form $\frac{1}{0}$, which indicates that the limit is either $\infty$, $-\infty$, or does not exist.

The term $(\sqrt{1+x^2} + \sqrt{1-x})$ approaches 2 as $x \to 0$, which is a positive value. The sign of the denominator is determined by the sign of $x$.

• As $x \to 0^+$ (x approaches 0 from the positive side), $x > 0$. The denominator $x(\sqrt{1+x^2} + \sqrt{1-x})$ approaches $0 \times 2 = 0$ from the positive side ($0^+$). So, $\lim_{x \to 0^+} \frac{x+1}{x(\sqrt{1+x^2} + \sqrt{1-x})} = \frac{1}{0^+} = +\infty$.

• As $x \to 0^-$ (x approaches 0 from the negative side), $x < 0$. The denominator $x(\sqrt{1+x^2} + \sqrt{1-x})$ approaches $0 \times 2 = 0$ from the negative side ($0^-$). So, $\lim_{x \to 0^-} \frac{x+1}{x(\sqrt{1+x^2} + \sqrt{1-x})} = \frac{1}{0^-} = -\infty$.

Since the left-hand limit ($-\infty$) and the right-hand limit ($+\infty$) are not equal, the limit does not exist.

The final answer is $\boxed{\text{The limit does not exist}}$.