Solution to Elementary Analysis: The Theory of Calculus Second Edition Section 9 Exercise 9.1

Solution:

### Part a

Let $s_n=\dfrac{n+1}{n}$. We have

$$

s_n=1+\frac{1}{n}.

$$ By Theorem 9.7(a) we have $\lim\dfrac{1}{n}=0$. Thus by Theorem 9.3, we have

$$

\lim \left(1+\frac{1}{n}\right)=\lim 1+\lim\frac{1}{n}=1+0=1.

$$

### Part b

Let $s_n=\dfrac{3n+7}{6n-5}$. We have

$$

s_n=\frac{3+\frac{7}{n}}{6-\frac{5}{n}}.

$$ By Theorem 9.7(a) we have $\lim\dfrac{1}{n}=0$. Thus by Theorems 9.2 and 9.3, we have

$$

\lim \left(3+\frac{7}{n}\right)=\lim 3+7\lim\frac{1}{n}=3+7\cdot 0=3,

$$ and

$$

\lim \left(6-\frac{5}{n}\right)=\lim 6-5\lim\frac{1}{n}=6-5\cdot 0=6.

$$ Then by Theorem 9.6 we obtain

$$

\lim \frac{3+\frac{7}{n}}{6-\frac{5}{n}}=\frac{3}{6}=\frac{1}{2}.

$$

### Part c

Let $s_n=\dfrac{17n^5+73n^4-18n^2+3}{23n^5+13n^3}$. We have

$$

s_n=\frac{17+\frac{73}{n}-\frac{18}{n^3}+\frac{3}{n^5}}{23+\frac{13}{n^2}}.

$$ By Theorem 9.7(a) we have

$$

\lim \frac{1}{n}=\lim\frac{1}{n^2}=\lim\frac{1}{n^3}=\lim\frac{1}{n^5}=0.

$$ Hence by Theorems 9.2 and 9.3 we have

\begin{align*}

&\ \lim\left(17+\frac{73}{n}-\frac{18}{n^3}+\frac{3}{n^5}\right)\\

=& \lim 17+73\lim\frac{1}{n}-18\frac{1}{n^3}+3\frac{1}{n^5}\\

=&\ 17+73\cdot 0-18\cdot 0+3\cdot 0= 17

\end{align*} and

$$

\lim\left(23+\frac{13}{n^2}\right)=23+13\lim\frac{1}{n^2}=23+13\cdot 0=23.

$$ Therefore by Theorem 9.6, we obtain

$$

\lim \dfrac{17n^5+73n^4-18n^2+3}{23n^5+13n^3}=\dfrac{17}{23}.

$$