Tính các giới hạn sau:
a) \(\lim\limits_{n\rightarrow+\infty}\)\(\left(\sqrt{n+2}-n\right)\)
b) \(\lim\limits_{n\rightarrow+\infty}\)\(\dfrac{2^n-3^{n-1}}{3^{n+2}}\)
c)\(\lim\limits_{n\rightarrow+\infty}\)\(\dfrac{\left(-1\right)^n}{5^{n-1}}\)
d)\(\lim\limits_{n\rightarrow+\infty}\)\(\dfrac{n\left(n+1\right).sin\left(n\right)}{n^2-2n+3}\)
a) \(\lim\limits_{n\rightarrow+\infty}\left(\sqrt{n+2}-n\right)\)\(=\lim\limits_{n\rightarrow+\infty}\dfrac{n+2-n^2}{\sqrt[]{n+2}+n}\)\(=-\infty\)
b) \(\lim\limits_{n\rightarrow+\infty}\dfrac{2^n-3^{n-1}}{3^{n+2}}\)\(=\lim\limits_{n\rightarrow+\infty}\dfrac{\left(\dfrac{2}{3}\right)^n-\dfrac{1}{3}}{9}\)\(=\dfrac{0-\dfrac{1}{3}}{9}=-\dfrac{1}{27}\)
c) \(\lim\limits_{n\rightarrow+\infty}\dfrac{\left(-1\right)^n}{5^{n-1}}\)\(=\lim\limits_{n\rightarrow+\infty}5.\left(\dfrac{-1}{5}\right)^n=5.0=0\)
d) \(\lim\limits_{n\rightarrow+\infty}\dfrac{n\left(n+1\right).sin\left(n\right)}{n^2-2n+3}\)\(=\lim\limits_{n\rightarrow+\infty}\dfrac{n^2\left(1+\dfrac{1}{n}\right).sin\left(n\right)}{n^2\left(1-\dfrac{2}{n}+\dfrac{3}{n^2}\right)}\)\(=\lim\limits_{n\rightarrow+\infty}\dfrac{\left(1+\dfrac{1}{n}\right).sin\left(n\right)}{\left(1-\dfrac{2}{n}+\dfrac{3}{n^2}\right)}\)
\(=1.\lim\limits_{n\rightarrow+\infty}sin\left(n\right)=\)\(\lim\limits_{n\rightarrow+\infty}sin\left(n\right)\)
vì \(-1\le sin\left(n\right)\le1\)
Nên theo định lý kẹp giới hạn trên không tồn tại.
