Bài tự luận · Bài 10
Dạng 1. Xác định giới hạn của dãy số
Bài tự luận Giới hạn của dãy số · Bài 10
(CTST11) Tìm các giới hạn sau:
a) $\lim \frac{(2n-1)(2n+3)}{2n^2 +4}$;
b) $\lim \frac{4n+1}{\sqrt{n^2 +3n}+n}$;
c) $\lim \sqrt{n}(\sqrt{n+1}-\sqrt{n})$;
d) $\lim \frac{1}{\sqrt{n^2 +n}-n}$.
a) $\lim \frac{(2n-1)(2n+3)}{2n^2 +4}$;
b) $\lim \frac{4n+1}{\sqrt{n^2 +3n}+n}$;
c) $\lim \sqrt{n}(\sqrt{n+1}-\sqrt{n})$;
d) $\lim \frac{1}{\sqrt{n^2 +n}-n}$.
Xem lời giải
Lời giải
a) $\lim \frac{(2n-1)(2n+3)}{2n^2 +4}=\lim \frac{\left( 2-\frac{1}{n} \right)\left( 2+\frac{3}{n} \right)}{2+\frac{4}{n^2}}=\frac{2.2}{2}=2$;
b) $\lim \frac{4n+1}{\sqrt{n^2 +3n}+n}=\lim \frac{4+\frac{1}{n}}{\sqrt{1+\frac{3}{n}}+1}=\frac{4+\lim \frac{1}{n}}{\sqrt{1+\lim \frac{3}{n}}+1}=\frac{4}{1+1}=2$;
c) $\lim \sqrt{n}(\sqrt{n+1}-\sqrt{n})=\lim \frac{\sqrt{n}(\sqrt{n+1}-\sqrt{n})(\sqrt{n+1}+\sqrt{n})}{\sqrt{n+1}+\sqrt{n}}=\lim \frac{\sqrt{n}}{\sqrt{n+1}+\sqrt{n}}$
$=\lim \frac{1}{\sqrt{1+\frac{1}{n}}+1}=\frac{1}{\sqrt{1+\lim \frac{1}{n}}+1}=\frac{1}{2}$;
d)
$\begin{array}{ll} \lim \frac{1}{\sqrt{n^2 +n}-n} & =\lim \frac{\sqrt{n^2 +n}+n}{\left( \sqrt{n^2 +n}-n \right)\left( \sqrt{n^2 +n}+n \right)}=\lim \frac{\sqrt{n^2 +n}+n}{n} \\ & =\lim \left( \sqrt{1+\frac{1}{n}}+1 \right)=2. \\ \end{array}$
b) $\lim \frac{4n+1}{\sqrt{n^2 +3n}+n}=\lim \frac{4+\frac{1}{n}}{\sqrt{1+\frac{3}{n}}+1}=\frac{4+\lim \frac{1}{n}}{\sqrt{1+\lim \frac{3}{n}}+1}=\frac{4}{1+1}=2$;
c) $\lim \sqrt{n}(\sqrt{n+1}-\sqrt{n})=\lim \frac{\sqrt{n}(\sqrt{n+1}-\sqrt{n})(\sqrt{n+1}+\sqrt{n})}{\sqrt{n+1}+\sqrt{n}}=\lim \frac{\sqrt{n}}{\sqrt{n+1}+\sqrt{n}}$
$=\lim \frac{1}{\sqrt{1+\frac{1}{n}}+1}=\frac{1}{\sqrt{1+\lim \frac{1}{n}}+1}=\frac{1}{2}$;
d)
$\begin{array}{ll} \lim \frac{1}{\sqrt{n^2 +n}-n} & =\lim \frac{\sqrt{n^2 +n}+n}{\left( \sqrt{n^2 +n}-n \right)\left( \sqrt{n^2 +n}+n \right)}=\lim \frac{\sqrt{n^2 +n}+n}{n} \\ & =\lim \left( \sqrt{1+\frac{1}{n}}+1 \right)=2. \\ \end{array}$