# Evaluating $\lim_{x\to 0} \dfrac{\sqrt{1-x}-\sqrt{1+x}}{x^2-3x}$

$$\lim_{x\to 0} \dfrac{\sqrt{1-x}-\sqrt{1+x}}{x^2-3x}$$

I am stuck at radicals. Division by 1/x doesn't help.

• I formatted your post. Tell me if this was not the intended meaning. Oct 17, 2013 at 5:58
• The Maple command $$Student[Calculus1]:-LimitTutor((sqrt(1-x)-sqrt(1+x))/(x^2-3*x), x = 0)$$ finds it step by step with explanation. See that link for info. BTW, I have strong reasons to think that some users of MathematicsSE have many accounts and vote up themselves. Oct 17, 2013 at 7:16
• The Maple commands won't do your exams for you and, even worse, won't let you get into the marvelous world of thinking and reasoning to get (closer, at least) at some problem's solution. Oct 17, 2013 at 10:57

Rationalize the numerator $$\sqrt{1-x}-\sqrt{1+x}=\frac{(1-x)-(1+x)}{\sqrt{1-x}+\sqrt{1+x}}=\frac{-2x}{\sqrt{1-x}+\sqrt{1+x}}$$
Then cancel out $x$ form the numerator & denominator as $x\ne0$ as $x\to0$
• @RCola, we get $$\frac{-2x}{\sqrt{1-x}+\sqrt{1+x}}\cdot\frac1{x(x-3)}=\frac{-2}{(\sqrt{1-x}+\sqrt{1+x})(x-3)}$$ Put $x=0$ Oct 17, 2013 at 6:17