Bài tự luận · Bài 7
Dạng 1. Xác định giới hạn của dãy số
Bài tự luận Giới hạn của dãy số · Bài 7
(CTST11) Tìm các giới hạn sau:
a) $\lim \frac{n^2 -2n+1}{2-3n^2}$
b) $\lim \frac{2n^2 +n-3}{n^3 +5}$
c) $\lim \left( \sqrt{n^2 +2n}-n \right)$
a) $\lim \frac{n^2 -2n+1}{2-3n^2}$
b) $\lim \frac{2n^2 +n-3}{n^3 +5}$
c) $\lim \left( \sqrt{n^2 +2n}-n \right)$
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Lời giải
a) $\lim \frac{n^2 -2n+1}{2-3n^2}=\lim \frac{1-\frac{2}{n}+\frac{1}{n^2}}{\frac{2}{n^2}-3}=\frac{\lim 1-\lim \frac{2}{n}+\lim \frac{1}{n^2}}{\lim \frac{2}{n^2}-3}=\frac{1-0+0}{0-3}=-\frac{1}{3}$.
b) $\lim \frac{2n^2 +n-3}{n^3 +5}=\lim \frac{\frac{2}{n}+\frac{1}{n^2}-\frac{3}{n^3}}{1+\frac{5}{n^3}}=\frac{\lim \frac{2}{n}+\lim \frac{1}{n^2}-\lim \frac{3}{n^3}}{1+\lim \frac{5}{n^3}}=\frac{0+0-0}{1+0}=0$
c) $\lim \left( \sqrt{n^2 +2n}-n \right)=\lim \frac{\left( \sqrt{n^2 +2n}-n \right)\left( \sqrt{n^2 +2n}+n \right)}{\sqrt{n^2 +2n}+n}=\lim \frac{\left( n^2 +2n \right)-n^2}{\sqrt{n^2 +2n}+n}$
$=\lim \frac{2n}{\sqrt{n^2 \left( 1+\frac{2}{n} \right)}+n}=\lim \frac{2}{\sqrt{1+\frac{2}{n}}+1}=\frac{\lim 2}{\lim \left( \sqrt{1+\frac{2}{n}}+1 \right)} \\ =\frac{2}{\sqrt{\lim 1+\lim \frac{2}{n}}+\lim 1}=\frac{2}{\sqrt{1+\lim \frac{2}{n}}+1}=\frac{2}{\sqrt{1+0}+1}=1.$
b) $\lim \frac{2n^2 +n-3}{n^3 +5}=\lim \frac{\frac{2}{n}+\frac{1}{n^2}-\frac{3}{n^3}}{1+\frac{5}{n^3}}=\frac{\lim \frac{2}{n}+\lim \frac{1}{n^2}-\lim \frac{3}{n^3}}{1+\lim \frac{5}{n^3}}=\frac{0+0-0}{1+0}=0$
c) $\lim \left( \sqrt{n^2 +2n}-n \right)=\lim \frac{\left( \sqrt{n^2 +2n}-n \right)\left( \sqrt{n^2 +2n}+n \right)}{\sqrt{n^2 +2n}+n}=\lim \frac{\left( n^2 +2n \right)-n^2}{\sqrt{n^2 +2n}+n}$
$=\lim \frac{2n}{\sqrt{n^2 \left( 1+\frac{2}{n} \right)}+n}=\lim \frac{2}{\sqrt{1+\frac{2}{n}}+1}=\frac{\lim 2}{\lim \left( \sqrt{1+\frac{2}{n}}+1 \right)} \\ =\frac{2}{\sqrt{\lim 1+\lim \frac{2}{n}}+\lim 1}=\frac{2}{\sqrt{1+\lim \frac{2}{n}}+1}=\frac{2}{\sqrt{1+0}+1}=1.$