Solution to Elementary Analysis: The Theory of Calculus Second Edition Section 7 Exercise 7.5

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}.

$$ Now it is clear that $\lim_{n\to 0} s_n=0$.

### Part b

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

\begin{align*}

s_n

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

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

=&\ \frac{1}{\sqrt{1+\frac{1}{n}}+1}.

\end{align*} In the last step, we simultaneously divide numerator and denominantor by $n$. Now it is clear that $\lim_{n\to 0} s_n=\dfrac{1}{2}$.

### Part c

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

\begin{align*}

s_n

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

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

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

\end{align*} In the last step, we simultaneously divide numerator and denominantor by $n$. Now it is clear that $\lim_{n\to 0} s_n=\dfrac{1}{4}$.