Đáp án:
$\lim\dfrac{(n^2 +1)^3(2n-1)^4}{4n^{10}-5}=4$
Giải thích các bước giải:
5d) $\lim\dfrac{(n^2 +1)^3(2n-1)^4}{4n^{10}-5}$
$=\lim\dfrac{n^6\left(1 +\dfrac{1}{n^2}\right)^3.n^4\left(2 -\dfrac1n\right)^4}{n^{10}\left(4 -\dfrac{5}{n^{10}}\right)}$
$=\lim\dfrac{\left(1 +\dfrac{1}{n^2}\right)^3\left(2 -\dfrac1n\right)^4}{4 -\dfrac{5}{n^{10}}}$
$=\dfrac{(1+0)^3(2-0)^4}{4 - 0}$
$=\dfrac{2^4}{4}$
$= 4$