$\lim_{n \to 1} (3n^4 -n -2)/(3n^2 – 4n +1) 12/07/2021 Bởi Reese $\lim_{n \to 1} (3n^4 -n -2)/(3n^2 – 4n +1)
Lời giải: $lim_{n->1}\frac{3n^4-n-2}{3n^2-4n+1}$ $=lim_{n->1}\frac{(3n^3+3n^2+3n+2)(n-1)}{(3n-1)(n-1)}$ $=lim_{n->1}\frac{3n^3+3n^2+3n+2}{3n-1}$ $=lim_{n->1}\frac{3n^3.(1+\frac{1}{n}+\frac{1}{n^2}+\frac{2}{3n^3})}{3n.(1-\frac{1}{3n})}$ $=\frac{3n^3}{3n}$ $=n^2$ $=1^2$ $=1$ Bình luận
Lời giải:
$lim_{n->1}\frac{3n^4-n-2}{3n^2-4n+1}$
$=lim_{n->1}\frac{(3n^3+3n^2+3n+2)(n-1)}{(3n-1)(n-1)}$
$=lim_{n->1}\frac{3n^3+3n^2+3n+2}{3n-1}$
$=lim_{n->1}\frac{3n^3.(1+\frac{1}{n}+\frac{1}{n^2}+\frac{2}{3n^3})}{3n.(1-\frac{1}{3n})}$
$=\frac{3n^3}{3n}$
$=n^2$
$=1^2$
$=1$
Đáp án:
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