lim x →1 ( √2x-1 + x ² -3x+1) / (∛(x-2) + x ²-x+1) Giúp mình với ạ. Cảm ơn ạ <3

Đáp án:

\[\mathop {\lim }\limits_{x \to 1} \frac{{\sqrt {2x – 1}  + {x^2} – 3x + 1}}{{\sqrt[3]{{x – 2}} + {x^2} – x + 1}} = 0\]

Giải thích các bước giải:

\(\begin{array}{l}
\mathop {\lim }\limits_{x \to 1} \frac{{\sqrt {2x – 1}  + {x^2} – 3x + 1}}{{\sqrt[3]{{x – 2}} + {x^2} – x + 1}}\\
 = \mathop {\lim }\limits_{x \to 1} \frac{{\left( {\sqrt {2x – 1}  – 1} \right) + {x^2} – 3x + 2}}{{\left( {\sqrt[3]{{x – 2}} + 1} \right) + \left( {{x^2} – x} \right)}}\\
 = \mathop {\lim }\limits_{x \to 1} \frac{{\frac{{2x – 1 – 1}}{{\sqrt {2x – 1}  + 1}} + \left( {x – 1} \right)\left( {x – 2} \right)}}{{\frac{{x – 2 + 1}}{{{{\sqrt[3]{{x – 2}}}^2} – \sqrt[3]{{x – 2}} + 1}} + x\left( {x – 1} \right)}}\\
 = \mathop {\lim }\limits_{x \to 1} \frac{{\frac{{2\left( {x – 1} \right)}}{{\sqrt {2x – 1}  + 1}} + \left( {x – 1} \right)\left( {x – 2} \right)}}{{\frac{{x – 1}}{{{{\sqrt[3]{{x – 2}}}^2} – \sqrt[3]{{x – 2}} + 1}} + x\left( {x – 1} \right)}}\\
 = \mathop {\lim }\limits_{x \to 1} \frac{{\frac{2}{{\sqrt {2x – 1}  + 1}} + x – 2}}{{\frac{1}{{{{\sqrt[3]{{x – 2}}}^2} – \sqrt[3]{{x – 2}} + 1}} + x}}\\
 = \frac{{\frac{2}{{\sqrt {2.1 – 1}  + 1}} + 1 – 2}}{{\frac{1}{{{{\sqrt[3]{{1 – 2}}}^2} – \sqrt[3]{{1 – 2}} + 1}} + 1}}\\
 = \frac{{\frac{2}{2} – 1}}{{\frac{1}{3} + 1}} = 0
\end{array}\)