Question

Iim √1+х - √1-х
x-o 3х

Answer 1

$\displaystyle \lim_{x \to 0}\dfrac{\sqrt{1+x}-\sqrt{1-x}}{3x}=\left\{\dfrac{0}{0}\right\}=\lim_{x \to 0}\dfrac{(\sqrt{1+x}-\sqrt{1-x})(\sqrt{1+x}+\sqrt{1-x})}{3x(\sqrt{1+x}+\sqrt{1-x})}=\\ \\ \\ =\lim_{x \to 0}\dfrac{\left(\sqrt{1+x}\right)^2-\left(\sqrt{1-x}\right)^2}{3x\left(\sqrt{1+x}+\sqrt{1-x}\right)}=\lim_{x \to 0}\dfrac{1+x-1+x}{3x\left(\sqrt{1+x}+\sqrt{1-x}\right)}=\\ \\ \\ =\lim_{x \to 0}\dfrac{2x}{3x\left(\sqrt{1+0}+\sqrt{1-0}\right)}=\dfrac{2}{3\cdot 2}=\dfrac{1}{3}$