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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?

user3195997
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  • What you did is a well known algebraic method to solve indeterminate forms, in particular when dealing with limits. Another such method used sometimes is l'Hospital's Rule. All is right and it means the limit of the first expresion when $;x;$ approaches zero is $;-1;$ – DonAntonio Feb 07 '16 at 12:40
  • Maybe I am missing some basic stuff, but what I mean is that if the first expression is $-1$ when $x$ approaches zero but not when $x$ is zero, then there is a difference between the first expression and the last one. So, why do we use equality in these two expressions if the result is not the same for every value of $x$? – user3195997 Feb 07 '16 at 12:52

1 Answers1

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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.

SchrodingersCat
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  • Thanks, it makes sense. But then is it correct to put an equal sign when making this kind of assumptions? – user3195997 Feb 07 '16 at 12:55
  • This is the exact explanation: if you don't assume $;x\neq0;$ then you cannot cancel $;x;$ in the last step, so the equality is justified only when $;x\neq 0;$, otherwise that's still an indeterminate form. – DonAntonio Feb 07 '16 at 12:57
  • @user3195997 Yes it is correct. But you should clearly mention the assumption. Just as Joanpemo said. – SchrodingersCat Feb 07 '16 at 13:00