Solution to Elementary Analysis: The Theory of Calculus Second Edition Section 8 Exercise 8.8

We shall use the following useful formula.

$\sqrt{a}-b=\dfrac{a-b^2}{\sqrt{a}+b}$, which is a variation of $(a-b)(a+b)=a^2-b^2$.

Solution:

### Part a

Using the formula $\sqrt{a}-b=\dfrac{a-b^2}{\sqrt{a}+b}$, We have

$$

s_n=\dfrac{(n^2+1)-n^2}{\sqrt{n^2+1}+n}=\frac{1}{\sqrt{n^2+1}+n}.

$$ Let $\epsilon>0$. Let $N=\dfrac{1}{\epsilon}$, then we have for $n>N$

\begin{align*}

|s_n-0|

=&\ \frac{1}{\sqrt{n^2+1}+n}\\

<&\ \frac{1}{n+n}=\frac{2}{n}\\

<&\ \frac{1}{n} <\frac{1}{N}=\epsilon

\end{align*} as desired. Hence $\lim s_n=0$.

### Part b

Using the formula $\sqrt{a}-b=\dfrac{a-b^2}{\sqrt{a}+b}$, We have

$$

\sqrt{n^2+n}-n=\dfrac{(n^2+n)-n^2}{\sqrt{n^2+n}+n}=\frac{n}{\sqrt{n^2+n}+n}

$$ and

\begin{align*}

s_n-\dfrac{1}{2}

=&\ \dfrac{(n^2+n)-n^2}{\sqrt{n^2+n}+n}-\frac{1}{2}\\

=&\ \frac{n}{\sqrt{n^2+n}+n}-\frac{1}{2}\\

=&\ \frac{n-\sqrt{n^2+n}}{2(\sqrt{n^2+n}+n)}\\

=&\ -\frac{n}{2(\sqrt{n^2+n}+n)^2}.

\end{align*} Let $\epsilon>0$. Let $N=\dfrac{1}{\epsilon}$, then we have for $n>N$

\begin{align*}

\left|s_n-\dfrac{1}{2}\right|

=&\ \frac{n}{2(\sqrt{n^2+n}+n)^2}\\

<&\ \frac{n}{2(n+n)^2}=\frac{1}{8n}\\

<&\ \frac{1}{n} <\frac{1}{N}=\epsilon

\end{align*} as desired. Hence $\lim s_n=\dfrac{1}{2}$.

### Part c

Using the formula $\sqrt{a}-b=\dfrac{a-b^2}{\sqrt{a}+b}$, We have

$$

\sqrt{4n^2+n}-2n=\dfrac{(4n^2+n)-(2n)^2}{\sqrt{4n^2+n}+2n}=\frac{n}{\sqrt{4n^2+n}+2n}

$$ and

\begin{align*}

s_n-\dfrac{1}{4}

=&\ \frac{n}{\sqrt{4n^2+n}+2n}-\frac{1}{4}\\

=&\ \frac{2n-\sqrt{4n^2+n}}{4(\sqrt{4n^2+n}+2n)}\\

=&\ -\frac{n}{4(\sqrt{4n^2+n}+2n)^2}.

\end{align*} Let $\epsilon>0$. Let $N=\dfrac{1}{\epsilon}$, then we have for $n>N$

\begin{align*}

\left|s_n-\dfrac{1}{4}\right|

=&\ \frac{n}{4(\sqrt{4n^2+n}+2n)^2}\\

<&\ \frac{n}{4(2n+2n)^2}=\frac{1}{64n}\\

<&\ \frac{1}{n} <\frac{1}{N}=\epsilon

\end{align*} as desired. Hence $\lim s_n=\dfrac{1}{4}$.