Giải thích các bước giải:
$\lim \dfrac{n^4(\sqrt{4n^3-n}-\sqrt{n^3})}{5-2n^5}$
$=\lim \dfrac{n^4\sqrt{n^3}(\sqrt{4-\dfrac{1}{n^2}}-1)}{5-2n^5}$
$=\lim \dfrac{n^5\sqrt{n}(\sqrt{4-\dfrac{1}{n^2}}-1)}{5-2n^5}$
$=\lim \dfrac{\sqrt{n}(\sqrt{4-\dfrac{1}{n^2}}-1)}{\dfrac{5}{n^5}-2}$
$=\dfrac{+\infty(\sqrt{4-0}-1)}{0-2}$
$=-\infty$