(1) 因\(\left|\dfrac{3n^2}{n^2-4}-3\right|=\left|\dfrac{3n^2}{n^2-4}\right|<\dfrac{1}{n-4}<\varepsilon\)，从而\(\forall \varepsilon>0\)，取\(N=\left[4+\dfrac{1}{\varepsilon}\right]\)，当\(n>N\)时，有\(\left|\dfrac{3n^2}{n^2-4}-3\right|<\varepsilon\)
(2)\(\left|\dfrac{\sin n}{n}\right|<\dfrac{1}{n}\).从而\(\forall\varepsilon>0\)，取\(N=\left[\dfrac{1}{\varepsilon}\right]\)，当\(n>N\)时，有\(\left|\dfrac{\sin n}{n}\right|<\varepsilon\)
(3)\(\sqrt[n]{1+n}=\left(\sqrt{1+n}\cdot \sqrt{1+n}\cdot1\cdot1\cdots 1\right)^{\frac{1}{n}}<\dfrac{2\sqrt{1+n}+(n-2)}{n}<1+\dfrac{2\sqrt{1+n}}{n}<1+\dfrac{2\sqrt{2}}{n}\)
从而取\(N=\left[\dfrac{2\sqrt2}{\varepsilon}\right]\)，则有\(\forall \varepsilon>0,\)存在\(N\)使得\(n>N,\left|\sqrt[n]{1+n}-1\right|<\varepsilon\)
(4)
因为\(a\)固定，则一定存在一个\(k>a,k\in\mathbb{N}\)，从而
因\(\dfrac{a}{k}<1\)则\(\left(\dfrac{a}{k}\right)^{n-k}\to 0,n\to+\infty\)，从而一定存在\(N\)，使得\(\left(\dfrac{a}{k}\right)^{n-k}<\dfrac{k!}{a^k}\varepsilon\)，则\(\forall\varepsilon>0,\exists N\)当\(n>N\)有\(\left|\dfrac{a^n}{n!}\right|<\varepsilon\)
2.因\(\lim\limits_{n\to \infty}a_n=a\)则由极限定义，取\(\sqrt{a}\varepsilon>0,\exists N\)当\(n>N\)时，一定有\(|a_n-a|<\sqrt{a}\varepsilon\).
则\(\left|\sqrt{a_n}-\sqrt{a}\right|=\left|\dfrac{a_n-a}{\sqrt{a_n}+\sqrt{a}}\right|<\left|\dfrac{a_n-a}{\sqrt{a}}\right|\)
3.因\(\lim\limits_{n\to \infty}a_n=a\)则由极限定义，\(\forall\varepsilon>0,\exists N,\)当\(n>N\)时，有\(|a_n-a|<\varepsilon\).而\(\left||a_n|-|a|\right|<\left|a_n-a\right|<\varepsilon\)
(1)转化为证明：\(\lim\limits_{n\to\infty}a^m_n=a^m\)(\(m\)是常数)
做因式分解：
\[a^m_n-a^m=(a_n-a)\left(a^0a_n^{m-1}+aa_{n}^{m-2}+\cdots+a^{m-1}a_n^0\right) \]因\(a_n\)极限存在，从而\(a_n\)有界限，则\(\left(a^0a_n^{m-1}+aa_{n}^{m-2}+\cdots+a^{m-1}a_n^0\right)\)是一个有界变量，记为\(M\)，则因\(\lim\limits_{n\to\infty}a_n=a\)，由极限的定义.取\(\dfrac{\varepsilon}{M}\)，\(\exists N\)当\(n>N\)有\(\left|a^m_n-a^m\right|<\varepsilon\)
(2)由题，取\(\varepsilon=\log_b\left(\dfrac{\varepsilon}{b^{a}}+1\right)\)，则\(\exists N\)当\(n>N\)时，有\(\left|a_n-a\right|<\log_b\left(\dfrac{\varepsilon}{b^{a}}+1\right)\).从而\(\left|b^{a_n}-b^a\right|=b^{a}\left|b^{a_n-a}-1\right|<b^{a}\cdot \dfrac{\varepsilon}{b^a}=\varepsilon\)
(3)由题取\(\varepsilon=ab^{\varepsilon}-a\)则\(\exists N,\)当\(n>N\)有\(\left|a_n-a\right|<ab^{\varepsilon}-a\)，即\(\left|\dfrac{a_n}{a}-1\right|<b^{\varepsilon}-1\)，即\(-b^{\varepsilon}<\dfrac{a_n}{a}<b^{\varepsilon}\).当\(b>1\)时，\(\left|\log_ba_n-\log_ba\right|=\left|\log_b\dfrac{a_n}{a}\right|<\left|\log_bb^\varepsilon\right|=\varepsilon\).当\(0<b<1\)时，\(\left|\log_ba_n-\log_ba\right|=\left|\log_b\dfrac{a_n}{a}\right|<\left|\log_b(-b^{\varepsilon})\right|=|-\varepsilon|=\varepsilon\).当\(b=1\)时，无需说明.
(4)结合，(2)，(3)即可
(5)由题\(\forall\varepsilon>0,\exists N>0\)当\(n>N\)时，有\(|a_n-a|<\varepsilon\)，从而$$\left|\sin a_n-\sin a\right|=\left|2\cos\dfrac{a_n+a}{2}\sin\dfrac{a_n-a}{2}\right|<2\cdot 1\cdot\dfrac{|a_n-a|}{2}=|a_n-a|<\varepsilon$$
5.要证：\(\lim\limits_{n\to\infty}\dfrac{\log_an}{n}=0\)，即证：\(\lim\limits_{n\to \infty}\log_a\sqrt[n]{n}=0\).因\(\lim\limits_{n\to \infty}\sqrt[n]{n}=1\)，取\(\varepsilon=a^{\varepsilon}+1\)，则存在\(n\)，当\(n>N\)时，有\(\left|\sqrt[n]{n}-1\right|<a^{\varepsilon}-1\)即\(\sqrt[n]{n}<a^{\varepsilon}\)
.从而\(\left|\log_a\sqrt[n]{n}\right|<\left|\log_aa^{\varepsilon}\right|=\varepsilon\)