Đáp án: $B$
Giải thích các bước giải:
$\lim\limits_{x\to +\infty}\dfrac{x^{2018}.\sqrt{4x^2+1} }{(2x+1)^{2019}}$
$=\lim\limits_{x\to +\infty}\dfrac{x^{2018}.x.\sqrt{4+\dfrac{1}{x^2}} }{ x^{2019}.\Big(2+\dfrac{1}{x}\Big)^{2019} }$
$=\lim\limits_{x\to +\infty}\dfrac{\sqrt{4+\dfrac{1}{x^2}}}{ \Big(2+\dfrac{1}{x}\Big)^{2019} }$
$=\dfrac{\sqrt4}{2^{2019}}$
$=\dfrac{1}{2^{2018}}$
