Question

$lim_{x to 1} frac{sqrt[3]{x}-1}{sqrt{2+x}-sqrt{3x}}$

Answer 1

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

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