Cho biết Lim {{x^2} - a x + a - 1/ x - 1 = 1

4/20

Cho biết \(\mathop {\lim }\limits_{x \to 1} \frac{{{x^2} - a\,x + a - 1}}{{x - 1}} = 1\,\,\left( {a \in \mathbb{R}} \right)\). Tính \(M = {a^2} + 2a\).

\(M = 8.\)

\(M = - 1.\)

\(M = 1.\)

\(M = 3.\)

Giải thích

Chọn D

Ta có: \(\mathop {\lim }\limits_{x \to 1} \frac{{{x^2} - a\,x + a - 1}}{{x - 1}} = \mathop {\lim }\limits_{x \to 1} \frac{{\left( {{x^2} - 1} \right) - a\,\left( {x - 1} \right)}}{{x - 1}} = \mathop {\lim }\limits_{x \to 1} \frac{{\left( {x + 1} \right)\left( {x - 1} \right) - a\,\left( {x - 1} \right)}}{{x - 1}}\)

\( = \mathop {\lim }\limits_{x \to 1} \frac{{\left( {x - 1} \right)\left( {x + 1 - a} \right)}}{{x - 1}} = \mathop {\lim }\limits_{x \to 1} \left( {x + 1 - a} \right) = 2 - a = 1\)

Suy ra \(2 - a = 1 \Leftrightarrow a = 1\).

Vậy \(M = {a^2} + 2a = {1^2} + 2.1 = 3\).