\lim \frac { 2 n } { n + 1 } = 2
ແກ້ສຳລັບ l
l=\frac{1}{Re(\frac{1}{n+1})Im(n)+Im(\frac{1}{n+1})Re(n)}
2Re(\frac{1}{n+1})Im(n)+2Im(\frac{1}{n+1})Re(n)\neq 0\text{ and }n\neq -1
ແບ່ງປັນ
ສໍາເນົາຄລິບ
\left(2Re(\frac{1}{n+1})Im(n)+2Im(\frac{1}{n+1})Re(n)\right)l=2
ສົມຜົນຢູ່ໃນຮູບແບບມາດຕະຖານ.
\frac{\left(2Re(\frac{1}{n+1})Im(n)+2Im(\frac{1}{n+1})Re(n)\right)l}{2Re(\frac{1}{n+1})Im(n)+2Im(\frac{1}{n+1})Re(n)}=\frac{2}{2Re(\frac{1}{n+1})Im(n)+2Im(\frac{1}{n+1})Re(n)}
ຫານທັງສອງຂ້າງດ້ວຍ 2Re(n)Im(\left(n+1\right)^{-1})+2Im(n)Re(\left(n+1\right)^{-1}).
l=\frac{2}{2Re(\frac{1}{n+1})Im(n)+2Im(\frac{1}{n+1})Re(n)}
ການຫານດ້ວຍ 2Re(n)Im(\left(n+1\right)^{-1})+2Im(n)Re(\left(n+1\right)^{-1}) ຈະຍົກເລີກການຄູນດ້ວຍ 2Re(n)Im(\left(n+1\right)^{-1})+2Im(n)Re(\left(n+1\right)^{-1}).
l=\frac{1}{Re(\frac{1}{n+1})Im(n)+Im(\frac{1}{n+1})Re(n)}
ຫານ 2 ດ້ວຍ 2Re(n)Im(\left(n+1\right)^{-1})+2Im(n)Re(\left(n+1\right)^{-1}).
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