Đáp án:
$=+\infty$
Giải thích các bước giải:
$\lim\limits_{x\to \dfrac{1}{2}} \dfrac{8x^2-1}{6x^2-5x+1}\\=\lim\limits_{x\to \dfrac{1}{2}} \dfrac{x^2(8-\dfrac{1}{x^2})}{x^2(6-\dfrac{5}{x}+\dfrac{1}{x^2})}\\=\lim\limits_{x\to \dfrac{1}{2}} \dfrac{(8-\dfrac{1}{x^2})}{(6-\dfrac{5}{x}+\dfrac{1}{x^2})}\\=\lim\limits_{x\to \dfrac{1}{2}} \dfrac{(8-\dfrac{1}{(\dfrac{1}{2})^2})}{(6-\dfrac{5}{\dfrac{1}{2}}+\dfrac{1}{(\dfrac{1}{2})^2})}\\=\dfrac{4}{0}=+\infty$
