Đáp án:

$13)\dfrac{2}{3}\\ 14)\dfrac{3}{2}\\ 15)2\\ 16)1$

Giải thích các bước giải:

$13)\\ \displaystyle\lim_{x \to 1} \dfrac{\sqrt[3]{x}-1}{\sqrt{x}-1}\\ =\displaystyle\lim_{x \to 1} \dfrac{\left(\sqrt[3]{x}-1\right)\left(x+\sqrt{x}+1\right)}{(\sqrt{x}-1)\left(x+\sqrt{x}+1\right)}\\ =\displaystyle\lim_{x \to 1} \dfrac{x-1}{(\sqrt{x}-1)\left(x+\sqrt{x}+1\right)}\\ =\displaystyle\lim_{x \to 1} \dfrac{(\sqrt{x}-1)(\sqrt{x}+1)}{(\sqrt{x}-1)\left(x+\sqrt{x}+1\right)}\\ =\displaystyle\lim_{x \to 1} \dfrac{\sqrt{x}+1}{x+\sqrt{x}+1}\\ =\dfrac{\sqrt{1}+1}{1+\sqrt{1}+1}\\ =\dfrac{2}{3}\\ 14)\\ \displaystyle\lim_{x \to 1} \dfrac{\sqrt{2x-1}-\sqrt{x}}{1-\sqrt[3]{2-x}}\\ =\displaystyle\lim_{x \to 1} \dfrac{\left(\sqrt{2x-1}-\sqrt{x}\right)\left(\sqrt[3]{2-x}^2+\sqrt[3]{2-x}+1\right)}{\left(1-\sqrt[3]{2-x}\right)\left(\sqrt[3]{2-x}^2+\sqrt[3]{2-x}+1\right)}\\ =\displaystyle\lim_{x \to 1} \dfrac{\left(\sqrt{2x-1}-\sqrt{x}\right)\left(\sqrt[3]{2-x}^2+\sqrt[3]{2-x}+1\right)}{\left(1-\sqrt[3]{2-x}\right)\left(\sqrt[3]{2-x}^2+\sqrt[3]{2-x}+1\right)}\\ =\displaystyle\lim_{x \to 1} \dfrac{\left(\sqrt{2x-1}-\sqrt{x}\right)\left(\sqrt[3]{2-x}^2+\sqrt[3]{2-x}+1\right)}{1-(2-x)}\\ =\displaystyle\lim_{x \to 1} \dfrac{\left(\sqrt{2x-1}-\sqrt{x}\right)\left(\sqrt[3]{2-x}^2+\sqrt[3]{2-x}+1\right)}{x-1}\\ =\displaystyle\lim_{x \to 1} \dfrac{\left(\sqrt{2x-1}-\sqrt{x}\right)\left(\sqrt{2x-1}+\sqrt{x}\right)\left(\sqrt[3]{2-x}^2+\sqrt[3]{2-x}+1\right)}{(x-1)\left(\sqrt{2x-1}+\sqrt{x}\right)}\\ =\displaystyle\lim_{x \to 1} \dfrac{(x-1)\left(\sqrt[3]{2-x}^2+\sqrt[3]{2-x}+1\right)}{(x-1)\left(\sqrt{2x-1}+\sqrt{x}\right)}\\ =\displaystyle\lim_{x \to 1} \dfrac{\sqrt[3]{2-x}^2+\sqrt[3]{2-x}+1}{\sqrt{2x-1}+\sqrt{x}}\\ =\dfrac{\sqrt[3]{2-1}^2+\sqrt[3]{2-1}+1}{\sqrt{2-1}+\sqrt{1}}\\ =\dfrac{3}{2}\\ 15)\displaystyle\lim_{x \to 2} \dfrac{\sqrt{3x-2}-\sqrt{x+2}}{\sqrt[3]{3x+2}-2}\\ =\displaystyle\lim_{x \to 2} \dfrac{\left(\sqrt{3x-2}-\sqrt{x+2}\right)\left(\sqrt{3x-2}+\sqrt{x+2}\right)\left(\sqrt[3]{3x+2}^2+2\sqrt[3]{3x+2}+4\right)}{\left(\sqrt[3]{3x+2}-2\right)\left(\sqrt[3]{3x+2}^2+2\sqrt[3]{3x+2}+4\right)\left(\sqrt{3x-2}+\sqrt{x+2}\right)}\\ =\displaystyle\lim_{x \to 2} \dfrac{(3x-2-x-2)\left(\sqrt[3]{3x+2}^2+2\sqrt[3]{3x+2}+4\right)}{(3x+2-8)\left(\sqrt{3x-2}+\sqrt{x+2}\right)}\\ =\displaystyle\lim_{x \to 2} \dfrac{(2x-4)\left(\sqrt[3]{3x+2}^2+2\sqrt[3]{3x+2}+4\right)}{(3x-6)\left(\sqrt{3x-2}+\sqrt{x+2}\right)}\\ =\dfrac{2}{3}\displaystyle\lim_{x \to 2} \dfrac{\sqrt[3]{3x+2}^2+2\sqrt[3]{3x+2}+4}{\sqrt{3x-2}+\sqrt{x+2}}\\ =\dfrac{2}{3}.\dfrac{\sqrt[3]{3.2+2}^2+2\sqrt[3]{3.2+2}+4}{\sqrt{3.2-2}+\sqrt{.2+2}}\\=2\\ 16)\\ \displaystyle\lim_{x \to 0} \dfrac{\sqrt{1+x^m}-\sqrt{1-x^m}}{x^m}\\ =\displaystyle\lim_{x \to 0} \dfrac{\left(\sqrt{1+x^m}-\sqrt{1-x^m}\right)\left(\sqrt{1+x^m}+\sqrt{1-x^m}\right)}{x^m\left(\sqrt{1+x^m}+\sqrt{1-x^m}\right)}\\ =\displaystyle\lim_{x \to 0} \dfrac{1+x^m-1+x^m}{x^m\left(\sqrt{1+x^m}+\sqrt{1-x^m}\right)}\\ =\displaystyle\lim_{x \to 0} \dfrac{2x^m}{x^m\left(\sqrt{1+x^m}+\sqrt{1-x^m}\right)}\\ =\displaystyle\lim_{x \to 0} \dfrac{2}{\sqrt{1+x^m}+\sqrt{1-x^m}}\\ =\dfrac{2}{\sqrt{1+0^m}+\sqrt{1-0^m}}\\ =\dfrac{2}{2}\\ =1$