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## Prove the limits equal to the corresponding numbers

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 $\lim s_n=\dfrac{1}{4}$.