Question

-1/3 limita (x- >0) radical din[(2-x^2)^3-2√2)]/x^2.

Answer 1

[tex]-\frac{1}{3} \lim_{x \to 0}\frac{{\sqrt{2-x^2}^3-2\sqrt{2}}}{x^2}= -\frac{1}{3} \lim_{x \to 0} \frac{(\sqrt{2-x^2}-\sqrt2)(\sqrt{2-x^2}^2+\sqrt2\sqrt{2-x^2}+2)}{x^2}=\\ -\frac{1}{3} \lim_{x \to 0}\frac{(2-x^2-2)(\sqrt{2-x^2}^2+\sqrt2\sqrt{2-x^2}+2)}{x^2(\sqrt{2-x^2}+\sqrt2)}=\frac{1}{3}\frac{6}{2\sqrt2}=\frac{\sqrt2}{2} [/tex]