Giải thích các bước giải:
$\lim_{x\to 1}\dfrac{\sqrt{2x-1}+x^2-3x+1}{\sqrt[3]{x-2}+x^2-x+1}$
$=\lim_{x\to 1}\dfrac{\sqrt{2x-1}-1+x^2-3x+2}{\sqrt[3]{x-2}+1+x^2-x}$
$=\lim_{x\to 1}\dfrac{\dfrac{2x-1-1}{\sqrt{2x-1}+1}+x^2-3x+2}{\dfrac{x-2+1}{\sqrt[3]{x-2}^2-\sqrt[3]{x-2}+1}+x^2-x}$
$=\lim_{x\to 1}\dfrac{\dfrac{2(x-1)}{\sqrt{2x-1}+1}+(x-1)(x-2)}{\dfrac{x-1}{\sqrt[3]{x-2}^2-\sqrt[3]{x-2}+1}+x(x-1)}$
$=\lim_{x\to 1}\dfrac{\dfrac{2}{\sqrt{2x-1}+1}+x-2}{\dfrac{1}{\sqrt[3]{x-2}^2-\sqrt[3]{x-2}+1}+x}$
$=\dfrac{\dfrac{2}{\sqrt{2-1}+1}+1-2}{\dfrac{1}{\sqrt[3]{1-2}^2-\sqrt[3]{1-2}+1}+1}$
$=0$
