1. Tìm lim [1 / (5n+3)] 2 Tìm lim [1 / (2n+7)] 05/11/2021 Bởi Bella 1. Tìm lim [1 / (5n+3)] 2 Tìm lim [1 / (2n+7)]
1. $\lim\dfrac{1}{5n+3}$ $=\lim\dfrac{ \dfrac{1}{n}}{5+\dfrac{3}{n}}$ $=\dfrac{0}{5+0}$ $=0$ 2. $\lim\dfrac{1}{2n+7}$ $=\lim\dfrac{ \dfrac{1}{n}}{2+\dfrac{7}{n}}$ $=\dfrac{0}{2+0}$ $=0$ Bình luận
1. lim$\frac{1}{5n+3}$ = lim$\frac{1}{n(5+\frac{3}{n}) }$ = lim$\frac{ \frac{1}{n} }{5+\frac{3}{n} }$ = $\frac{0}{5+0}$ = 0 2. lim$\frac{1}{2n+7}$ = lim$\frac{1}{n(2+\frac{7}{n}) }$ = lim$\frac{ \frac{1}{n} }{2+\frac{7}{n} }$ = $\frac{0}{2+0}$ = 0 Bình luận
1.
$\lim\dfrac{1}{5n+3}$
$=\lim\dfrac{ \dfrac{1}{n}}{5+\dfrac{3}{n}}$
$=\dfrac{0}{5+0}$
$=0$
2.
$\lim\dfrac{1}{2n+7}$
$=\lim\dfrac{ \dfrac{1}{n}}{2+\dfrac{7}{n}}$
$=\dfrac{0}{2+0}$
$=0$
1. lim$\frac{1}{5n+3}$
= lim$\frac{1}{n(5+\frac{3}{n}) }$
= lim$\frac{ \frac{1}{n} }{5+\frac{3}{n} }$
= $\frac{0}{5+0}$
= 0
2. lim$\frac{1}{2n+7}$
= lim$\frac{1}{n(2+\frac{7}{n}) }$
= lim$\frac{ \frac{1}{n} }{2+\frac{7}{n} }$
= $\frac{0}{2+0}$
= 0