Evaluate the limit: lim (x->0) [sqrt(1+x) - sqrt(1-x)] / x

Substitute $x=0$ into the expression:

$$\frac{\sqrt{1+0} - \sqrt{1-0}}{0} = \frac{\sqrt{1} - \sqrt{1}}{0} = \frac{1-1}{0} = \frac{0}{0}$$

This is an indeterminate form, so we need to simplify the expression.

Multiply the numerator and the denominator by $(\sqrt{1+x} + \sqrt{1-x})$:

$$\lim_{x \to 0} \frac{\sqrt{1+x} - \sqrt{1-x}}{x} \times \frac{\sqrt{1+x} + \sqrt{1-x}}{\sqrt{1+x} + \sqrt{1-x}}$$

Here, $a = \sqrt{1+x}$ and $b = \sqrt{1-x}$.

$$\lim_{x \to 0} \frac{(\sqrt{1+x})^2 - (\sqrt{1-x})^2}{x(\sqrt{1+x} + \sqrt{1-x})}$$ $$\lim_{x \to 0} \frac{(1+x) - (1-x)}{x(\sqrt{1+x} + \sqrt{1-x})}$$ $$\lim_{x \to 0} \frac{1+x-1+x}{x(\sqrt{1+x} + \sqrt{1-x})}$$ $$\lim_{x \to 0} \frac{2x}{x(\sqrt{1+x} + \sqrt{1-x})}$$

Since $x \to 0$, $x \neq 0$, so we can cancel $x$ from the numerator and denominator:

$$\lim_{x \to 0} \frac{2}{\sqrt{1+x} + \sqrt{1-x}}$$

$$\frac{2}{\sqrt{1+0} + \sqrt{1-0}} = \frac{2}{\sqrt{1} + \sqrt{1}} = \frac{2}{1+1} = \frac{2}{2} = 1$$

The limit is 1.