lim x^2 + 2x +1 / 2x^3 + 2 bằng

Ta có \(f\left( x \right) = \frac{{{x^2} + 2x + 1}}{{2{x^3} + 2}} = \frac{{{{\left( {x + 1} \right)}^2}}}{{2\left( {x + 1} \right)\left( {{x^2} - x + 1} \right)}} = \frac{{x + 1}}{{2\left( {{x^2} - x + 1} \right)}}\)

Suy ra \[\mathop {\lim }\limits_{x \to  - 1} \frac{{{x^2} + 2x + 1}}{{2{x^3} + 2}} = \mathop {\lim }\limits_{x \to  - 1} f\left( x \right) = \mathop {\lim }\limits_{x \to  - 1} \frac{{x + 1}}{{2\left( {{x^2} - x + 1} \right)}} = \frac{0}{6} = 0.\]

Đáp án: 0.