giải em câu này đi ạ lim √n^2+n – n^2+n/√4n^4+n – 2n^2 03/07/2021 Bởi Cora giải em câu này đi ạ lim √n^2+n – n^2+n/√4n^4+n – 2n^2
Giải thích các bước giải: $\lim\dfrac{\sqrt{n^2+n}-n^2+n}{\sqrt{4n^4+n}-2n^2}$ $=\lim\dfrac{\sqrt{n^2+n}-n^2+n}{\dfrac{4n^4+n-4n^4}{\sqrt{4n^4+n}+2n^2}}$ $=\lim\dfrac{\sqrt{n^2+n}-n^2+n}{\dfrac{n}{\sqrt{4n^4+n}+2n^2}}$ $=\lim\dfrac{(\sqrt{n^2+n}-n^2+n)(\sqrt{4n^4+n}+2n^2)}{n}$ $=\lim n^3.\dfrac{(\sqrt{n^2+n}-n^2+n)(\sqrt{4n^4+n}+2n^2)}{n^4}$ $=\lim n^3.(\sqrt{\dfrac{1}{n^2}+\dfrac{1}{n^3}}-1+\dfrac{1}{n})(\sqrt{4+\dfrac{1}{n}}+2)$ $=+\infty.(\sqrt{0+0}-1+0)(\sqrt{4+0}+2)$ $=-\infty$ Bình luận
Giải thích các bước giải:
$\lim\dfrac{\sqrt{n^2+n}-n^2+n}{\sqrt{4n^4+n}-2n^2}$
$=\lim\dfrac{\sqrt{n^2+n}-n^2+n}{\dfrac{4n^4+n-4n^4}{\sqrt{4n^4+n}+2n^2}}$
$=\lim\dfrac{\sqrt{n^2+n}-n^2+n}{\dfrac{n}{\sqrt{4n^4+n}+2n^2}}$
$=\lim\dfrac{(\sqrt{n^2+n}-n^2+n)(\sqrt{4n^4+n}+2n^2)}{n}$
$=\lim n^3.\dfrac{(\sqrt{n^2+n}-n^2+n)(\sqrt{4n^4+n}+2n^2)}{n^4}$
$=\lim n^3.(\sqrt{\dfrac{1}{n^2}+\dfrac{1}{n^3}}-1+\dfrac{1}{n})(\sqrt{4+\dfrac{1}{n}}+2)$
$=+\infty.(\sqrt{0+0}-1+0)(\sqrt{4+0}+2)$
$=-\infty$