Giải thích các bước giải:
$\lim_{x\to1}\dfrac{\sqrt{2x-1}-\sqrt[3]{x^2+x-1}}{x^3-2x^2+x}$
$=\lim_{x\to1}\dfrac{(\sqrt{2x-1}-x)-(\sqrt[3]{x^2+x-1}-x)}{x(x-1)^2}$
$=\lim_{x\to1}\dfrac{\dfrac{2x-1-x^2}{\sqrt{2x-1}+x}-\dfrac{x^2+x-1-x^3}{\sqrt[3]{x^2+x-1}^2+x\sqrt[3]{x^2+x-1}+x^2}}{x(x-1)^2}$
$=\lim_{x\to1}\dfrac{\dfrac{-(x-1)^2}{\sqrt{2x-1}+x}-\dfrac{-(x-1)^2(x+1)}{\sqrt[3]{x^2+x-1}^2+x\sqrt[3]{x^2+x-1}+x^2}}{x(x-1)^2}$
$=\lim_{x\to1}\dfrac{\dfrac{-(x-1)^2}{\sqrt{2x-1}+x}+\dfrac{(x-1)^2(x+1)}{\sqrt[3]{x^2+x-1}^2+x\sqrt[3]{x^2+x-1}+x^2}}{x(x-1)^2}$
$=\lim_{x\to1}\dfrac{\dfrac{-1}{\sqrt{2x-1}+x}+\dfrac{x+1}{\sqrt[3]{x^2+x-1}^2+x\sqrt[3]{x^2+x-1}+x^2}}{x}$
$=\dfrac{\dfrac{-1}{\sqrt{2.1-1}+1}+\dfrac{1+1}{\sqrt[3]{1^2+1-1}^2+1.\sqrt[3]{1^2+1-1}+1^2}}{1}$
$=\dfrac16$
