Indeterminate form and equality of expressions.

Given this chain of equalities:

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

Equality should imply that, for any value of $x$, the result in each side of the equation is going to be the same. But in this case, for $x=0$, we obtain $\frac{0}{0}$ in the first expression and $-1$ in the last one. What is happening here? Are all the transformations correct?

Answer

When you are cancelling $x$ in the $2$nd last step from both the numerator and denominator, you are making the assumption that $x\not = 0$. If $x=0$, then you cannot cancel the $x$'s from numerator and denominator.