Hola, como calcular el lìmite de la siguiente funciòn cuando x tiende a cero ?

Candy maritza!

$$\begin{align}&\lim_{x \to 0} \frac{\sqrt{2-x}- \sqrt{2+x}}{x^2+x}=\frac{0}{0}=\\&\\&\lim_{x \to 0} \frac{(\sqrt{2-x}- \sqrt{2+x})(\sqrt{2-x}+ \sqrt{2+x})}{(x^2+x)(\sqrt{2-x}+\sqrt{2+x})}=\\&\\&\lim_{x \to 0} \frac{(\sqrt{2-x})^2- (\sqrt{2+x})^2}{(x^2+x)(\sqrt{2-x}+\sqrt{2+x})}=\\&\\&\lim_{x \to 0} \frac{2-x- (2+x)}{(x^2+x)(\sqrt{2-x}+\sqrt{2+x})}=\\&\\&\lim_{x \to 0} \frac{-2x}{x(x+1)(\sqrt{2-x}+\sqrt{2+x})}=\\&\\&\lim_{x \to 0} \frac{-2}{(x+1)(\sqrt{2-x}+\sqrt{2+x})}=\\&\\&\frac{-2}{\sqrt 2+\sqrt 2}=\frac{-2}{2 \sqrt 2}=\frac{-1}{\sqrt 2}=\\&\\&\frac{-1}{\sqrt 2}\frac{ \sqrt 2}{ \sqrt 2}=\frac{-\sqrt 2}{2}\end{align}$$