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Formal proof of limits I

Solution:

Part a

Let $\epsilon>0$ and let $N=\dfrac{1}{\epsilon}$. Then $n>N$ implies
$$\frac{1}{n}<\frac{1}{N}=\epsilon,$$ and hence
$$\left|\dfrac{(-1)^n}{n}-0\right|=\frac{1}{n}<\epsilon$$ as desired. Therefore $\lim\dfrac{(-1)^n}{n}=0$.

Part b

Let $\epsilon>0$ and let $N=\dfrac{1}{\epsilon^3}$. Then $n>N$ implies
$$\frac{1}{n^{1/3}}<\dfrac{1}{N^{1/3}}=\frac{1}{1/\epsilon}=\epsilon.$$ and hence
$$\left|\frac{1}{n^{1/3}}-0\right|=\frac{1}{n^{1/3}}<\epsilon.$$ as desired. Therefore $\lim\dfrac{1}{n^{1/3}}=0$.

Part c

Let $\epsilon>0$ and let $N=\dfrac{7}{9\epsilon}$. Then $n>N$ implies
$$\frac{7}{9n+6}<\frac{7}{9N}=\epsilon.$$ and hence
\begin{align*}
\left|\frac{2n-1}{3n+2}-\frac{2}{3}\right|=\frac{7}{9n+6}<\epsilon.
\end{align*} as desired. Therefore $\lim\dfrac{2n-1}{3n+2}=\dfrac{2}{3}$.

Part d

Let $\epsilon>0$ and let $N=6+\dfrac{1}{\epsilon}$. Then $n>N$ implies
$$0<\frac{n+6}{n^2-6}<\frac{n+6}{n^2-36}=\frac{1}{n-6}<\frac{1}{N-6}=\epsilon,$$ and hence
$$\left|\frac{n+6}{n^2-6}-0\right|=\frac{n+6}{n^2-6}<\epsilon$$ as desired. Therefore $\lim\dfrac{n+6}{n^2-6}=0$.