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Determine the limits and prove your claims


Solution:

Part a

The limit is zero.

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


Part b

The limit is $\dfrac{7}{3}$.

Let $\epsilon>0$ and let $N=\dfrac{106}{9\epsilon}$. Then $n>N$ implies
$$
\frac{106}{9n+21}<\frac{106}{9N}=\epsilon.
$$ and hence
\begin{align*}
\left|\frac{7n-19}{3n+7}-\frac{7}{3}\right|=\frac{106}{9n+21}<\epsilon.
\end{align*} as desired. Therefore $\lim\dfrac{7n-19}{3n+7}=\dfrac{7}{3}$.


Part c

The limit is $\dfrac{4}{7}$.

Let $\epsilon>0$ and let $N=\dfrac{41}{49\epsilon}+\dfrac{5}{7}$. Then $n>N$ implies
$$
\frac{41}{49n-35}<\frac{41}{49(\frac{41}{49\epsilon}+\frac{5}{7})-35}=\epsilon.
$$ and hence
\begin{align*}
\left|\frac{4n+3}{7n-5}-\frac{4}{7}\right|=\frac{41}{49n-35}<\epsilon.
\end{align*} as desired. Therefore $\lim\dfrac{4n+3}{7n-5}=\dfrac{4}{7}$.


Part d

The limit is $\dfrac{2}{5}$.

Let $\epsilon>0$ and let $N=\dfrac{16}{25\epsilon}$. Then $n>N$ implies
$$
\frac{16}{25n+10}<\frac{16}{25n}<\frac{16}{25N}=\epsilon,
$$ and hence
$$
\left|\frac{2n+4}{5n+2}-\dfrac{2}{5}\right|=\frac{16}{25n+10}<\epsilon
$$ as desired. Therefore $\lim\dfrac{2n+4}{5n+2}=\dfrac{2}{5}$.


Part e

The limit is $0$.

Let $\epsilon>0$ and let $N=\dfrac{1}{\epsilon}$. Then $n>N$ implies
$$
\left|\frac{1}{n}\sin n\right| \le \frac{1}{n}<\frac{1}{N}=\epsilon,
$$ here we used $|\sin n|\le 1$, and hence
$$
\left|\frac{1}{n}\sin n-0\right|=\left|\frac{1}{n}\sin n\right|<\epsilon
$$ as desired. Therefore $\lim \dfrac{1}{n}\sin n=0$.


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