Toán 1. Tìm (n^2 + n) / (12n^2 + 1) 2. Tìm (5n + 3) / (2n + 1) 05/11/2021 By Margaret 1. Tìm (n^2 + n) / (12n^2 + 1) 2. Tìm (5n + 3) / (2n + 1)
1. $\lim\dfrac{n^2+n}{12n^2+1}$ $=\lim\dfrac{1+\dfrac{1}{n}}{12+\dfrac{1}{n^2}}$ $=\dfrac{1}{12}$ 2. $\lim\dfrac{5n+3}{2n+1}$ $=\lim\dfrac{5+\dfrac{3}{n}}{2+\dfrac{1}{n}}$ $=\dfrac{5}{2}$ Trả lời
1) lim$\frac{n²+n}{12n²+1}$ = lim$\frac{n²(1+1/n)}{n²(12+1/n²)}$ = lim$\frac{1+1/n}{12+1/n²}$ = $\frac{1}{12}$ 2) lim$\frac{5n+3}{2n+1}$ = lim$\frac{n(5+3/n)}{n(2+1/n)}$ = lim$\frac{5+3/n}{2+1/n}$ = $\frac{5}{2}$ Trả lời
1.
$\lim\dfrac{n^2+n}{12n^2+1}$
$=\lim\dfrac{1+\dfrac{1}{n}}{12+\dfrac{1}{n^2}}$
$=\dfrac{1}{12}$
2.
$\lim\dfrac{5n+3}{2n+1}$
$=\lim\dfrac{5+\dfrac{3}{n}}{2+\dfrac{1}{n}}$
$=\dfrac{5}{2}$
1) lim$\frac{n²+n}{12n²+1}$
= lim$\frac{n²(1+1/n)}{n²(12+1/n²)}$
= lim$\frac{1+1/n}{12+1/n²}$
= $\frac{1}{12}$
2) lim$\frac{5n+3}{2n+1}$
= lim$\frac{n(5+3/n)}{n(2+1/n)}$
= lim$\frac{5+3/n}{2+1/n}$
= $\frac{5}{2}$