Đáp án: $a.\lim\dfrac{4n(n^2+1)}{(2n+4)^3}=\dfrac12$
$e.\lim_{x\to+\infty}\dfrac{3|x|+\sqrt{4x^2+x}}{x+1}=5$
Giải thích các bước giải:
$a.\lim\dfrac{4n(n^2+1)}{(2n+4)^3}$
$=\lim\dfrac{4(1+\dfrac1{n^2})}{(2+\dfrac4n)^3}$
$=\dfrac{4(1+0)}{(2+0)^3}$
$=\dfrac12$
$e.\lim_{x\to+\infty}\dfrac{3|x|+\sqrt{4x^2+x}}{x+1}$
$=\lim_{x\to+\infty}\dfrac{3x+\sqrt{4x^2+x}}{x+1}$
$=\lim_{x\to+\infty}\dfrac{3+\sqrt{4+\dfrac1x}}{1+\dfrac1x}$
$=\dfrac{3+\sqrt{4+0}}{1+0}$
$=5$