\input homework_macros.tex
\def\Li{\displaystyle\lim_{n\to\infty}}
%\distiller
\printer

\def\dd#1#2{\frac{d#1}{d#2}}
\toks0={
8.2  p. 628:   5,  7, 11, 15, 27, 37, 59.
}

\SN{\jobname}
\np
$a_n=\Frac{1-5n^4}{n^4+8n^3}$.
Either use L'H\^opital's Rule four times or rewrite
$a_n=\Frac{\frac 1{n^4}-5}{1+8\frac{n^3}{n^4}}=
\Frac{\frac 1{n^4}-5}{1+8\frac{1}{n}}=
\Frac{0-5}{1+8\cdot 0}=-5$.
Since we have computed the limit, the sequence converges.

\np
$a_n=\Frac{n^2-2n+1}{n-1}$.
You can use L'H\^opital's Rule or else
$a_n=\Frac{n^2-2n+1}{n-1}=\Frac{(n-1)^2}{n-1}=n-1$
so $\Li a_n=\Li n-1=\infty$ so the sequence diverges.
Another way to rewrite $a_n$ is
$a_n=n\cdot\Frac{\frac{n^2}{n^2}-2\cdot\frac{n}{n^2}+\frac 1{n^2}}
{\frac nn-\frac 1n}=
n\cdot\Frac{1-2\cdot\frac1{n}+\frac1{n^2}}{1-\frac1n}$.
Now $\Li\Frac{1-2\cdot\frac1{n}+\frac1{n^2}}{1-\frac1n}=1$ so
$\Li a_n=(\Li n)\cdot
\Bigl(\Li\Frac{1-2\cdot\frac1{n}+\frac1{n^2}}{1-\frac1n}\Bigr)=
\infty\cdot 1=\infty$.

\np
$a_n=\Bigl(\Frac{n+1}{2n}\Bigr)\Bigl(1-\Frac 1n\Bigr)$.
$\Li a_n=\Bigl(\Li \Frac{n+1}{2n}\Bigr)\Bigl(\Li 1-\Frac 1n\Bigr)=
\frac 12\cdot 1=\frac12$.
This computation shows that the sequence converges to $\frac12$.

\np
$a_n=\sqrt{\Frac{2n}{n+1}\ }$.
First do any easier computation:
$\Li\Frac{2n}{n+1}=\Li\Frac{2}{1+\frac 1n}=2$.
Since the square root function is continuous at 2,
the sequence converges and
$\Li\sqrt{\Frac{2n}{n+1}\ }=\sqrt{\Frac{2n}{n+1}\ }=\sqrt 2$
(see Theorem 4).

\np
$a_n=\Bigl(1+\frac 7n \Bigr)^n$.
Study $f(x)=\Bigl(1+\frac 7x \Bigr)^x$ for $x>0$ and
compute $\displaystyle\lim_{x\to\infty} f(x)$.
This has the form $1^\infty$ which is indeterminate.
Take the log:
$\ln f(x)= x\ln\bigl(1+\frac 7x\bigr)$.
$\displaystyle\lim_{x\to\infty}\ln f(x)$ has the form $\infty\cdot \ln1=
\infty\cdot0$, which is again indeterminate.
Hence rewrite
$x\ln\bigl(1+\frac 7x\bigr)=\Frac{\ln\bigl(1+\frac 7x\bigr)}{\frac1x}$.
To compute 
$\displaystyle\lim_{x\to\infty}\Frac{\ln\bigl(1+\frac 7x\bigr)}{\frac1x}$
use L'H\^opital's Rule.
$\displaystyle\lim_{x\to\infty}\Frac{\ln\bigl(1+\frac 7x\bigr)}{\frac1x}=
\displaystyle\lim_{x\to\infty}
\Frac{\frac{-\frac7{x^2}}{1+\frac7x}}{-\frac1{x^2}}=
\displaystyle\lim_{x\to\infty}\Frac 7{1+\frac7x}=7$.
Unravelling these calculations, we see
$\displaystyle\lim_{x\to\infty}\ln f(x)=7$, so
$\displaystyle\lim_{x\to\infty}f(x)=e^7$ and hence the
sequence converges and
$\Li a_n=e^7$.

\np
$a_n=\Frac{n!}{n^n}$.
(Hint: compare with $\frac 1n$.)
Well, $\Frac{n!}{n^n}=\frac 1n\cdot\frac2n\cdots\frac{n-1}n\cdot\frac nn=
\frac1n\cdot\bigl(\frac2n\cdots\frac{n-1}n\cdot\frac nn)$.
The factor $\frac2n\cdots\frac{n-1}n\cdot\frac nn$ is a product of
positive fractions, each of which is less than $1$ so
$0<\frac2n\cdots\frac{n-1}n\cdot\frac nn<1$ and hence
$0<a_n<\frac 1n$.
Since $\Li \frac1n=0$ the Sandwich Theorem shows
that the sequence converges and $\Li a_n=0$.

\vfill\eject
\np
$a_n=n-\sqrt{n^2-n\ }$.
Study the function $f(x)=x-\sqrt{x^2-x\ }$ and in praticular the
limit $\displaystyle\lim_{x\to\infty}f(x)$.
This has the form $\infty-\infty$ so L'H\^opital's Rule techniques
can be used.
Write $f(x)=x\cdot\bigl(1-\sqrt{1-\frac1x\ }\bigr)=
\Frac{1-\sqrt{1-\frac1x\ }}{\frac 1x}$.
This last expression has the form $\frac 00$ as $x\to\infty$ so
L'H\^opital's Rule can be applied.
$\displaystyle\lim_{x\to\infty}\Frac{1-\sqrt{1-\frac1x\ }}{\frac 1x}=
\displaystyle\lim_{x\to\infty}
\Frac{0-\Biggl(\Frac{-\bigl(-\frac1{x^2}\bigr)}
{2\sqrt{1-\frac1x}}\Biggr)}{-\frac 1{x^2}}=
\displaystyle\lim_{x\to\infty}\Frac{1}{2\sqrt{1-\frac1x}}=
\Frac{1}{2\sqrt{1-0}}=\frac12$.
Hence the sequence converges and $\Li a_n=\frac12$.

\end