$1)$
Câu 1:
$S=12+\dfrac{1}{3^0}+\dfrac{1}{3^2}+\dfrac{1}{3^{n-3}}+...$
$=12+\dfrac{1}{1-\dfrac{1}{3}}$
$=\dfrac{27}{2}$
Câu 2:
$S=\sqrt2\Big( \dfrac{1}{2^0}+\dfrac{1}{2^1}+\dfrac{1}{2^2}+...+\dfrac{1}{2^n}...\Big)$
$=\sqrt2.\dfrac{1}{1-\dfrac{1}{2}}$
$=2\sqrt2$
$2)$
1.
$\lim\dfrac{2n^2-3}{-2n^3-4}=\lim\dfrac{ \dfrac{-2}{n}-\dfrac{3}{n^3} }{-2-\dfrac{4}{n^3}}=0$
2.
$\lim\dfrac{n^3-2n}{5n^2+n-3}$
$=\lim\dfrac{n^3\Big(1-\dfrac{2}{n^2}\Big)}{n^2\Big(5-\dfrac{1}{n}-\dfrac{3}{n^2}\Big)}$
$=\lim n.\dfrac{1-\dfrac{2}{n^2}}{5-\dfrac{1}{n}-\dfrac{3}{n^2}}$
$=+\infty$
3.
$\lim(\sqrt{n^2-3n+1}+n)$
$=\lim n(\sqrt{1-\dfrac{3}{n}+\dfrac{1}{n^2}}+1)$
$=+\infty$
4.
$\lim(\sqrt{n^2-3n+1}-n)$
$=\lim\dfrac{n^2-3n+1-n^2}{\sqrt{n^2-3n+1}+n}$
$=\lim\dfrac{-3n+1}{\sqrt{n^2-3n+1}+n}$
$=\lim\dfrac{-3+\dfrac{1}{n}}{\sqrt{1-\dfrac{3}{n}+\dfrac{1}{n^2}}+1}$
$=\dfrac{-3}{2}$
5.
$\lim(\sqrt{n^2-3n+1}-2n)$
$=\lim n(\sqrt{1-\dfrac{3}{n}+\dfrac{1}{n^2}}-2)$
$=-\infty$
