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

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