\(a=lim\frac{\left(\frac{2}{3}\right)^n+1}{3\left(\frac{1}{3}\right)^n-12}=-\frac{1}{12}\)

\(b=lim\frac{4\left(\frac{4}{10}\right)^n+1}{\left(\frac{3}{10}\right)^n-40}=-\frac{1}{40}\)

\(c=lim\frac{1-\left(\frac{2}{12}\right)^n}{1+45\left(\frac{3}{12}\right)^n}=\frac{1}{1}=1\)

\(d=\frac{\left(-\frac{2}{3}\right)^n+1}{-2\left(-\frac{2}{3}\right)^n-12+2\left(\frac{1}{3}\right)^n}=-\frac{1}{12}\)

\(e=\frac{1-11\left(\frac{1}{3}\right)^n}{\left(\frac{1}{3}\right)^n+14\left(\frac{2}{3}\right)^n}=\frac{1}{0}=+\infty\)

\(f=\frac{\left(\frac{2}{5}\right)^n-3+\left(\frac{1}{5}\right)^n}{3\left(\frac{2}{5}\right)^n+28\left(\frac{4}{5}\right)^n}=\frac{-3}{0}=-\infty\)

a) \(\lim\limits3=3\) vì \(3\) là hằng số.

Áp dụng giới hạn cơ bản với \(k=2\), ta có:\(\lim\limits\dfrac{1}{n^2}=0\).

b) \(\lim\limits\left(3+\dfrac{1}{n^2}\right)=\lim\limits3+\lim\limits\dfrac{1}{n^2}=3\).

a/ \(=lim\frac{1}{\sqrt{n+1}+\sqrt{n}}=\frac{1}{\infty}=0\)

b/ \(=lim\frac{6n+1}{\sqrt{n^2+5n+1}+\sqrt{n^2-n}}=\frac{6+\frac{1}{n}}{\sqrt{1+\frac{5}{n}+\frac{1}{n^2}}+\sqrt{1-\frac{1}{n}}}=\frac{6}{1+1}=3\)

c/ \(=lim\frac{6n-9}{\sqrt{3n^2+2n-1}+\sqrt{3n^2-4n+8}}=lim\frac{6-\frac{9}{n}}{\sqrt{3+\frac{2}{n}-\frac{1}{n^2}}+\sqrt{3-\frac{4}{n}+\frac{8}{n^2}}}=\frac{6}{\sqrt{3}+\sqrt{3}}=\sqrt{3}\)

d/ \(=lim\frac{\left(\frac{2}{6}\right)^n+1-4\left(\frac{4}{6}\right)^n}{\left(\frac{3}{6}\right)^n+6}=\frac{1}{6}\)

e/ \(=lim\frac{\left(\frac{3}{5}\right)^n-\left(\frac{4}{5}\right)^n+1}{\left(\frac{3}{5}\right)^n+\left(\frac{4}{5}\right)^n-1}=\frac{1}{-1}=-1\)

f/ Ta có công thức:

\(1+3+...+\left(2n+1\right)^2=\left(n+1\right)^2\)

\(\Rightarrow lim\frac{1+3+...+2n+1}{3n^2+4}=lim\frac{\left(n+1\right)^2}{3n^2+4}=lim\frac{\left(1+\frac{1}{n}\right)^2}{3+\frac{4}{n^2}}=\frac{1}{3}\)

g/ \(=lim\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{n}-\frac{1}{n+1}\right)=lim\left(1-\frac{1}{n+1}\right)=1-0=1\)

h/ Ta có: \(1^2+2^2+...+n^2=\frac{n\left(n+1\right)\left(2n+1\right)}{6}\)

\(\Rightarrow lim\frac{n\left(n+1\right)\left(2n+1\right)}{6n\left(n+1\right)\left(n+2\right)}=lim\frac{2n+1}{6n+12}=lim\frac{2+\frac{1}{n}}{6+\frac{12}{n}}=\frac{2}{6}=\frac{1}{3}\)

Đặt \(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+....+\frac{1}{n\left(n+1\right)}=A\)

\(\Leftrightarrow A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{n}-\frac{1}{n+1}\)

\(\Leftrightarrow A=\frac{n+1}{n+1}-\frac{1}{n+1}=\frac{n}{n+1}\)