Lim ( căn bậc 3 của 2x+1 ) – 1 )/x-1 khi x tiến tới 1

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1. Đáp án:

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

   Giải thích các bước giải:

    Ta có:

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

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