Answer 1

$\lim_{x \to 1} \frac{\sqrt[3]{x}-1}{\sqrt{2+x}-\sqrt{3x}}=\\ \lim_{x \to 1} \frac{(\sqrt[3]{x}-1)(\sqrt[3]{x^2}+\sqrt[3]{x}+1)(\sqrt{2+x}+\sqrt{3x})}{(\sqrt[3]{x^2}+\sqrt[3]{x}+1)(\sqrt{2+x}-\sqrt{3x})(\sqrt{2+x}+\sqrt{3x})}=\\ \lim_{x \to 1} \frac{(x-1)(\sqrt{2+x}+\sqrt{3x})}{(\sqrt[3]{x^2}+\sqrt[3]{x}+1)(2+x-3x)}=\\ \lim_{x \to 1} \frac{(x-1)(\sqrt{2+x}+\sqrt{3x})}{(\sqrt[3]{x^2}+\sqrt[3]{x}+1)(2-2x)}=\\ \lim_{x \to 1} \frac{(x-1)(\sqrt{2+x}+\sqrt{3x})}{-2(\sqrt[3]{x^2}+\sqrt[3]{x}+1)(x-1)}=$

$\lim_{x \to 1} \frac{\sqrt{2+x}+\sqrt{3x}}{-2(\sqrt[3]{x^2}+\sqrt[3]{x}+1)}=\\ \frac{\sqrt{2+1}+\sqrt{3*1}}{-2(\sqrt[3]{1^2}+\sqrt[3]{1}+1)}=\\ \frac{2\sqrt{3}}{-2*3}=\\ -\frac{\sqrt{3}}{3}$