Cho đa thức f(x) thỏa mãn \[\frac{{f\left( x \right) - 2}}{{x - 1}} = 12\]. Tính \[\mathop {\lim }\limits_{x \to 1} \frac{{f\left( x \right) - 2}}{{\left( {{x^2} - 1} \right)\left[ {f\left( x \right) + 1} \right]}}\]

A.\[\mathop {\lim }\limits_{x \to {x_0}} \left[ {f\left( x \right) + g\left( x \right)} \right] = L\]

B. \[\mathop {\lim }\limits_{x \to {x_0}} \left[ {f\left( x \right) + g\left( x \right)} \right] = M\]

C. \[\mathop {\lim }\limits_{x \to {x_0}} \left[ {f\left( x \right) + g\left( x \right)} \right] = L - M\]

D. \[\mathop {\lim }\limits_{x \to {x_0}} \left[ {f\left( x \right) + g\left( x \right)} \right] = M + L\]

A.0.

B.1.

C.2.

D.3.

A.\[ + \infty .\]

B.2.

C.4.

D.\[ - \infty .\]

A.\[T = \frac{{12}}{{25}}.\]

B. \[T = \frac{4}{{25}}.\]

C. \[T = \frac{4}{{15}}.\]

D. \[T = \frac{6}{{25}}.\]

A.\[\mathop {\lim }\limits_{x \to + \infty } \frac{{{x^2} + 1}}{{2{x^2} + 1}} = \frac{1}{2}\]

B. \[\mathop {\lim }\limits_{x \to - \infty } \left( {{x^2} + 3x - 1} \right) = - \infty \]

C. \[\mathop {\lim }\limits_{x \to + \infty } \frac{{x + 1}}{{2x + 1}} = \frac{1}{2}\]

D. \[\mathop {\lim }\limits_{x \to - \infty } \frac{{x + 3}}{{2x + 1}} = \frac{1}{2}\]

A.\[\mathop {\lim }\limits_{x \to + \infty } f\left( x \right) = L\]

B. \[\mathop {\lim }\limits_{x \to - \infty } f\left( x \right) = L\]

C. \[\mathop {\lim }\limits_{x \to {x_0}} f\left( x \right) = L\]

D. \[\mathop {\lim }\limits_{x \to L} f\left( x \right) = {x_0}\]

A.0.

B.\[ + \infty .\]

C. \[\sqrt 2 - 1.\]

D. \[ - \infty .\]