Bài tự luận · Bài 20
Dạng 1. Xác định giới hạn của dãy số
Bài tự luận Giới hạn của dãy số · Bài 20
(KNTT11) Tính các giới hạn sau:
a) $\lim\limits_{n\to+\infty} \left( \sqrt{n^2 +2n}-n-2 \right)$;
b) $\lim\limits_{n\to+\infty} \left( 2+n^2 -\sqrt{n^4 +1} \right)$;
c) $\lim\limits_{n\to+\infty} \left( \sqrt{n^2 -n+2}+n \right)$
d) $\lim\limits_{n\to+\infty} \left( 3n-\sqrt{4n^2 +1} \right)$.
a) $\lim\limits_{n\to+\infty} \left( \sqrt{n^2 +2n}-n-2 \right)$;
b) $\lim\limits_{n\to+\infty} \left( 2+n^2 -\sqrt{n^4 +1} \right)$;
c) $\lim\limits_{n\to+\infty} \left( \sqrt{n^2 -n+2}+n \right)$
d) $\lim\limits_{n\to+\infty} \left( 3n-\sqrt{4n^2 +1} \right)$.
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Lời giải
a) $\lim\limits_{n\to+\infty} \left( \sqrt{n^2 +2n}-n-2 \right)=\lim\limits_{n\to+\infty} \frac{-2n-4}{\left( \sqrt{n^2 +2n}+n+2 \right)}$ $=\lim\limits_{n\to+\infty} \frac{-2-\frac{4}{n}}{\left( \sqrt{1+\frac{2}{n}}+1+\frac{2}{n} \right)}=-1.$
b) $\lim\limits_{n\to+\infty} \left( 2+n^2 -\sqrt{n^4 +1} \right)=\lim\limits_{n\to+\infty} \frac{4n^2 +3}{2+n^2 +\sqrt{n^4 +1}}=\lim\limits_{n\to+\infty} \frac{4+\frac{3}{n^2}}{\frac{2}{n^2}+1+\sqrt{1+\frac{1}{n^4}}}=2$
c) $\lim\limits_{n\to+\infty} \left( \sqrt{n^2 -n+2}+n \right)=\lim\limits_{n\to+\infty} n\left( \sqrt{1-\frac{1}{n}+\frac{2}{n^2}}+1 \right)=+\infty$.
d) $\lim\limits_{n\to+\infty} \left( 3n-\sqrt{4n^2 +1} \right)=\lim\limits_{n\to+\infty} n\left( 3-\sqrt{4+\frac{1}{n^2}} \right)=+\infty$.
b) $\lim\limits_{n\to+\infty} \left( 2+n^2 -\sqrt{n^4 +1} \right)=\lim\limits_{n\to+\infty} \frac{4n^2 +3}{2+n^2 +\sqrt{n^4 +1}}=\lim\limits_{n\to+\infty} \frac{4+\frac{3}{n^2}}{\frac{2}{n^2}+1+\sqrt{1+\frac{1}{n^4}}}=2$
c) $\lim\limits_{n\to+\infty} \left( \sqrt{n^2 -n+2}+n \right)=\lim\limits_{n\to+\infty} n\left( \sqrt{1-\frac{1}{n}+\frac{2}{n^2}}+1 \right)=+\infty$.
d) $\lim\limits_{n\to+\infty} \left( 3n-\sqrt{4n^2 +1} \right)=\lim\limits_{n\to+\infty} n\left( 3-\sqrt{4+\frac{1}{n^2}} \right)=+\infty$.