2 câu trả lời
Đáp án:
$\lim \dfrac{\:4n^2+3n+1}{\left(3n-1\right)^2}$ $=\dfrac{4}{9}$
Giải thích các bước giải:
$\lim \dfrac{\:4n^2+3n+1}{\left(3n-1\right)^2}$
$=\lim \dfrac{n^2\left(4+\dfrac{3}{n}+\dfrac{1}{n^2}\right)}{\left(n\left(3-\dfrac{1}{n}\right)\right)^2}$
$=\lim \dfrac{4+\dfrac{3}{n}+\dfrac{1}{n^2}}{\left(3-\dfrac{1}{n}\right)^2}$
$=\dfrac{\lim 4+\dfrac{3}{n}+\dfrac{1}{n^2}}{\lim \left(3-\dfrac{1}{n}\right)^2}$
$=\dfrac{4}{9}$
Đáp án:
`lim\frac{4n^{2}+3n+1}{(3n-1)^{2}}=\frac{4}{9}`
Giải thích các bước giải:
`lim\frac{4n^{2}+3n+1}{(3n-1)^{2}}``=lim\frac{4n^{2}+3n+1}{9n^{2}-6n+1}``=lim\frac{n^{2}(4+\frac{3}{n}+\frac{1}{n^{2}})}{n^{2}(9-\frac{6}{n}+\frac{1}{n^{2}})}``=lim\frac{4+\frac{3}{n}+\frac{1}{n^{2}}}{9-\frac{6}{n}+\frac{1}{n^{2}}}=\frac{4}{9}`
