Giới hạn

D

- Hướng dẫn giải

Đáp án D

Ta có $\frac{1}{3x^2-4x-4}+\frac{1}{x^2-12x+20}=\frac{1}{\left(x-1\right)\left(3x+2\right)}+\frac{1}{\left(x-2\right)\left(x-10\right)}$

          $=\frac{x-10+3x+2}{\left(x-2\right)\left(3x+2\right)\left(x-10\right)}=\frac{4\left(x-2\right)}{\left(x-2\right)\left(3x+2\right)\left(x-10\right)}=\frac{4}{\left(3x+2\right)\left(x-10\right)}$ 

Do đó $\underset{x\to 2}{\lim }\left(\frac{1}{3x^2-4x-4}+\frac{1}{x^2-12x+20}\right)=\underset{x\to 2}{\lim }\frac{4}{\left(3x+2\right)\left(x-10\right)}=\frac{-1}{16}$

Vậy theo bài ra thì $a=-1,b=16$ nên $b-a=17$