The given problem is to evaluate the following limit:

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

Solution:

1. Multiply by the conjugate:
   To simplify the expression, multiply both the numerator and denominator by the conjugate of the numerator:

   $\frac{\sqrt{1+x} - \sqrt{1-x}}{x} \times \frac{\sqrt{1+x} + \sqrt{1-x}}{\sqrt{1+x} + \sqrt{1-x}} = \frac{\left(\sqrt{1+x} - \sqrt{1-x}\right)\left(\sqrt{1+x} + \sqrt{1-x}\right)}{x \left(\sqrt{1+x} + \sqrt{1-x}\right)}.$

2. Simplify the numerator:
   Applying the difference of squares formula $(a-b)(a+b) = a^2 - b^2$, the numerator simplifies as follows:

   $(\sqrt{1+x})^2 - (\sqrt{1-x})^2 = (1+x) - (1-x) = 1 + x - 1 + x = 2x.$

   Thus, the expression becomes:

   $\frac{2x}{x \left(\sqrt{1+x} + \sqrt{1-x}\right)}.$

3. Cancel out $x$:
   The $x$ in the numerator and denominator cancels out:

   $\frac{2}{\sqrt{1+x} + \sqrt{1-x}}.$

4. Evaluate the limit as $x \to 0$:
   Now substitute $x = 0$ into the simplified expression:

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

Thus, the value of the limit is:

$\boxed{1}.$

Would you like further details or have any questions?

Here are 5 related questions to deepen your understanding:

1. How does multiplying by the conjugate help in simplifying this limit?
2. What is the difference of squares formula and how is it applied here?
3. Why does the term $x$ cancel out in the simplified expression?
4. What are some other examples of limits where multiplying by the conjugate is useful?
5. How would the solution change if the limit involved $x \to 1$ instead of $x \to 0$?

Tip: When dealing with limits involving square roots, multiplying by the conjugate is a common technique to eliminate radicals.