Cho đa thức $\Large f(x)$ thỏa mãn $\Large \underset{x\rightarrow 2}{l

Cho đa thức $\Large f(x)$ thỏa mãn $\Large \underset{x\rightarrow 2}{lim}\dfrac{f(x)-20}{x-2}=10$. Tính $\Large T=\underset{x\rightarrow 2}{lim}\dfrac{\sqrt[3]{6f(x)+5}-5}{x^2+x-6}$.

• A.

  $\Large T=\dfrac{12}{25}$.

• B.

  $\Large T=+\infty$.

• C.

  $\Large T=\dfrac{4}{25}$.

• D.

  $\Large T=-\infty$.

Chọn C
Từ giả thiết $\Large \underset{x\rightarrow 2}{lim}\dfrac{f(x)-20}{x-2}=10$ $\Large \Rightarrow f(2)=20$.

$\Large T=\underset{x\rightarrow 2}{lim}\dfrac{\sqrt[3]{6f(x)+5}-5}{x^2+x-6}$ $\Large \underset{x\rightarrow 2}{lim}\dfrac{6f(x)-120}{(x-2)(x+3)\left[\sqrt[3]{6f(x)+5)^2}+5\sqrt[3]{6f(x)+5}+25\right]}$

$\Large \underset{x\rightarrow 2}{lim}\left[\dfrac{6\left(f(x)-20\right)}{(x-2)}.\dfrac{1}{(x+3)\left[\sqrt[3]{\left(6f(x)+5\right)^2}+5\sqrt[3]{6f(x)+5}+25\right]}\right]$ $\Large =60.\dfrac{1}{375}=\dfrac{4}{25}$.