Toán tìm giới hạn của hàm số Un= √(N ²+3N+2) -N+1 15/11/2021 By Cora tìm giới hạn của hàm số Un= √(N ²+3N+2) -N+1
$\lim(\sqrt{n^2+3n+2}-(n-1))$ $=\lim\dfrac{n^2+3n+2-(n-1)^2}{\sqrt{n^2+3n+2}+n-1}$ $=\lim\dfrac{n^2+3n+2-n^2+2n-1}{\sqrt{n^2+3n+2}+n-1}$ $=\lim\dfrac{5n+1}{\sqrt{n^2+3n+2}+n-1}$ $=\lim\dfrac{5n+1}{\sqrt{n^2\Big(1+\dfrac{3}{n}+\dfrac{2}{n^2}\Big)}+n-1}$ $=\lim\dfrac{5+\dfrac{1}{n}}{\sqrt{1+\dfrac{3}{n}+\dfrac{2}{n^2}}+1-\dfrac{1}{n}}$ $=\dfrac{5}{1+1}$ $=\dfrac{5}{2}$ Trả lời
Đáp án: $\lim U_n =\dfrac52$ Giải thích các bước giải: $\lim U_n =\lim(\sqrt{n^2 + 3n +2} – n +1)$ $=\lim\dfrac{(\sqrt{n^2 + 3n +2} – n +1)(\sqrt{n^2 + 3n +2} + n -1)}{\sqrt{n^2 + 3n +2} +n -1}$ $=\lim\dfrac{n^2 + 3n + 2 – (n-1)^2}{\sqrt{n^2 + 3n +2} +n -1}$ $=\lim\dfrac{5n +1}{\sqrt{n^2 + 3n +2} +n -1}$ $=\lim\dfrac{5 +\dfrac1n}{\sqrt{1+ \dfrac3n +\dfrac{2}{n^2}} + 1 -\dfrac1n}$ $=\dfrac{5+0}{\sqrt{1 + 3.0 + 2.0} + 1 – 0}$ $=\dfrac52$ Trả lời
$\lim(\sqrt{n^2+3n+2}-(n-1))$
$=\lim\dfrac{n^2+3n+2-(n-1)^2}{\sqrt{n^2+3n+2}+n-1}$
$=\lim\dfrac{n^2+3n+2-n^2+2n-1}{\sqrt{n^2+3n+2}+n-1}$
$=\lim\dfrac{5n+1}{\sqrt{n^2+3n+2}+n-1}$
$=\lim\dfrac{5n+1}{\sqrt{n^2\Big(1+\dfrac{3}{n}+\dfrac{2}{n^2}\Big)}+n-1}$
$=\lim\dfrac{5+\dfrac{1}{n}}{\sqrt{1+\dfrac{3}{n}+\dfrac{2}{n^2}}+1-\dfrac{1}{n}}$
$=\dfrac{5}{1+1}$
$=\dfrac{5}{2}$
Đáp án:
$\lim U_n =\dfrac52$
Giải thích các bước giải:
$\lim U_n =\lim(\sqrt{n^2 + 3n +2} – n +1)$
$=\lim\dfrac{(\sqrt{n^2 + 3n +2} – n +1)(\sqrt{n^2 + 3n +2} + n -1)}{\sqrt{n^2 + 3n +2} +n -1}$
$=\lim\dfrac{n^2 + 3n + 2 – (n-1)^2}{\sqrt{n^2 + 3n +2} +n -1}$
$=\lim\dfrac{5n +1}{\sqrt{n^2 + 3n +2} +n -1}$
$=\lim\dfrac{5 +\dfrac1n}{\sqrt{1+ \dfrac3n +\dfrac{2}{n^2}} + 1 -\dfrac1n}$
$=\dfrac{5+0}{\sqrt{1 + 3.0 + 2.0} + 1 – 0}$
$=\dfrac52$