Ta có
$\underset{x \to 1}{\lim} \dfrac{\sqrt{2x-1} + x^2 - 3x + 1}{\sqrt[3]{x-2} + x^2 - x + 1} = \underset{x \to 1}{\lim} \dfrac{(\sqrt{2x-1} - 1 + x^2 - 3x + 2}{\sqrt[3]{x-2} + 1 + x^2 - x}$
$= \underset{x \to 1}{\lim} \dfrac{\frac{2x-1-1}{\sqrt{2x-1} + 1} + (x-1)(x-2)}{\frac{x-2+1}{\sqrt[3]{(x-2)^2} + 1 - \sqrt[3]{x-2}} + x(x-1)}$
$= \underset{x \to 1}{\lim} \dfrac{\frac{2}{\sqrt{2x-1} + 1} + x-2}{\frac{1}{\sqrt[3]{(x-2)^2} + 1 - \sqrt[3]{x-2}} + x}$
$= \dfrac{\frac{2}{2} + 1 - 2}{\frac{1}{1 +1+1} + 1}$
$= 0$
Vậy
$\underset{x \to 1}{\lim} \dfrac{\sqrt{2x-1} + x^2 - 3x + 1}{\sqrt[3]{x-2} + x^2 - x + 1} = 0$
