\input homework_macros.tex

%\distiller
\printer
\def\Lim{\displaystyle\lim}

\def\dd#1#2{\frac{d#1}{d#2}}

\SN{6.6}
\pn7. For $\displaystyle\lim_{t\to0}\frac{\sin t^2}{t}$,
the form is $\frac 00$ so l'H\^opital's Rule applies.
First calculate $\dd{\sin t^2}{t}=-(\cos t^2)\dd{t^2}{t}=(-2t)\cos t^2$,
so our limit is
$\displaystyle\lim_{t\to0}\frac{(-2t)\cos t^2}{1}=
\lim_{t\to0}-(2t)\cos t^2=0$.

\pn13.  Find $\displaystyle\lim_{\theta\to\frac\pi2 }
\frac{1-\sin\theta}{1+\cos2\theta}$.
Since $\sin(\frac\pi2)=-1$ and $\cos(2\frac\pi2)=-1$,
the form is $\frac 00$.
Hence l'H\^opital's Rule applies and our limit is
$\displaystyle\lim_{\theta\to\frac\pi2 }
\frac{-\cos\theta}{-2\sin(2\theta)}$. 
Since $\cos(\frac\pi2)=0=\sin(2\frac\pi2)$, this second limit 
also has the form $\frac00$ and l'H\^opital's Rule applies.
$\displaystyle\lim_{\theta\to\frac\pi2 }
\frac{\sin\theta}{-4\cos(2\theta)}=\frac{1}{(-4)(-1)}=\frac 14$.

\pn15. Find
$\Lim_{x\to0}\frac{x^2}{\ln(\sec x)}$.
Since $\sec 0=1$, $\ln(\sec0)=0$ so we have the form $\frac00$
and l'H\^opital's Rule applies.
To apply l'H\^opital's Rule, we need to compute 
$\dd{\ln(\sec x)}x=\frac{\dd{\sec x}{x}}{\sec x}=
(\cos x)(\sec x)(\tan x)=\tan x$.
Our limit is
$\Lim_{x\to0}\frac{2x}{\tan x}$ which again has the form $\frac 00$.
There are at least two ways to proceed.
First apply l'H\^opital's Rule:
$\Lim_{x\to0}\frac{2}{\sec^2x}=\frac{2}{1}=2$.
A second way is to simplify first:
$\Lim_{x\to0}\frac{2x}{\tan x}=\Lim_{x\to0}(\cos x)\frac{2x}{\sin x}=
(2\cos x)\Lim_{x\to0}\frac{x}{\sin x}$ and we have calculated
$\Lim_{x\to0}\frac{x}{\sin x}=1$, so our limit is
$(2\cos(0))\cdot 1=2$.

\pn19. Find
$\Lim_{x\to\frac\pi2^-}\br2{x-\frac\pi2}\sec x$.
The form if $0\cdot\infty$ so we write 
$\Lim_{x\to\frac\pi2^-}\br2{x-\frac\pi2}\sec x=
\Lim_{x\to\frac\pi2^-}\frac{x-\frac\pi2}{\cos x}
$ and this limit has the form $\frac00$.
By l'H\^opital's Rule, 
$\Lim_{x\to\frac\pi2^-}\frac{x-\frac\pi2}{\cos x}=
\Lim_{x\to\frac\pi2^-}\frac{1}{-\sin x}$
and since $\sin(\frac\pi2)=1$, our limit is
$-1$.

\pn27. Find
$\Lim_{x\to0^+}\frac{\ln(x^2+2x)}{\ln x}$ which has the form
$\frac\infty\infty$.
To apply l'H\^opital's Rule, compute
$\dd{\ln(x^2+2x)}x=\frac{2x+2}{x^2+2x}$ so
$\Lim_{x\to0^+}\frac{\ln(x^2+2x)}{\ln x}=
\Lim_{x\to0^+}\frac{\frac{2x+2}{x^2+2x}}{\frac1x}=
\Lim_{x\to0^+}\frac{x(2x+2)}{x^2+2x}=
\Lim_{x\to0^+}\frac{2x+2}{x+2}=\frac 22=1$.

\pn31. Find $\Lim_{x\to\infty}\bigl(\ln(2x) -\ln(x+1)\bigr)$
which has the form $\infty-\infty$.
Write
$\Lim_{x\to\infty}\bigl(\ln(2x) -\ln(x+1)\bigr)=
\Lim_{x\to\infty} \bigl(\ln(x+1)\bigr)
\bigl(\frac{\ln(2x)}{\ln(x+1)}-1\bigr)$.
Compute $\Lim_{x\to\infty}\frac{\ln(2x)}{\ln(x+1)}$ using
l'H\^opital's Rule:
$\Lim_{x\to\infty}\frac{\ln(2x)}{\ln(x+1)}=
\Lim_{x\to\infty}\frac{\frac{2}{2x}}{\frac1{x+1}}=1$
so this time we have the form $\infty\cdot0$ and we rewrite again:
$\Lim_{x\to\infty}\bigl(\ln(2x) -\ln(x+1)\bigr)=
\Lim_{x\to\infty}\frac{\frac{\ln(2x)}{\ln(x+1)}-1}{\frac1{\ln(x+1)}}$
which has the form $\frac00$.
Compute $\dd{\frac{\ln(2x)}{\ln(x+1)}}x=
\frac{\dd{\ln(2x)}{x}\ln(x+1)-(\ln2x)\dd{\ln(x+1)}{x}}
{\bigl(\ln(x+1)\bigr)^2}=
\frac{\frac{\ln(x+1)}{x}-\frac{\ln(2x)}{x+1}}{\bigl(\ln(x+1)\bigr)^2}$;
$\dd{\frac1{\ln(x+1)}}{x}=-\frac1{(x+1)\bigl(\ln(x+1)\bigr)^2}$
so
$\Lim_{x\to\infty}\bigl(\ln(2x) -\ln(x+1)\bigr)=\Lim_{x\to\infty}
\frac{\frac{\frac{\ln(x+1)}x-\frac{\ln(2x)}{x+1}}{\bigl(\ln(x+1)\bigr)^2}}
{-\frac1{(x+1)\bigl(\ln(x+1)\bigr)^2}}=
-\Lim_{x\to\infty}\frac{(x+1)\ln(x+1)}{x}-\ln(2x)$.
This looks so much like the original problem that one feels
one is making no progress.
We try a different approach.
Note $e^{\Lim_{x\to\infty}\bigl(\ln(2x) -\ln(x+1)\bigr)}=
\Lim_{x\to\infty}e^{\ln(2x) -\ln(x+1)}=
\Lim_{x\to\infty}\frac{2x}{x+1}$.
This last limit can be computed using l'H\^opital's Rule again
to get $2$.
Since $e^{\Lim_{x\to\infty}\bigl(\ln(2x) -\ln(x+1)\bigr)}=2$,
$\Lim_{x\to\infty}\bigl(\ln(2x) -\ln(x+1)\bigr)=\ln2$.

\pn37.Find
$\Lim_{x\to\infty}\int_x^{2x}\frac 1t\>dt=\Lim{x\to\infty}
\ln(2x)-\ln x$.
Just as in 31, this last limit is $\ln2$.

\pn42.
$\Lim_{x\to\infty}x^2e^{-x}=\Lim_{x\to\infty}\frac{x^2}{e^x}$.
l'H\^opital's Rule applies ($\frac\infty\infty$) to get
$\Lim_{x\to\infty}\frac{2x}{e^x}$ which still has form 
$\frac\infty\infty$, so
$\Lim_{x\to\infty}\frac{2}{e^x}=0$.

\pn43. Find
$\Lim_{x\to1^+}x^{\frac1{1-x}}=\Lim_{x\to1^+}
e^{\frac{\ln x}{1-x}}=
e^{\Lim_{x\to1^+}\frac{\ln x}{1-x}}$.
We compute $\Lim_{x\to1^+}\frac{\ln x}{1-x}$
using l'H\^opital's Rule, so compute
$\Lim_{x\to1^+}\frac{\frac 1x}{-x}=-1$.
Hence $\Lim_{x\to1^+}x^{\frac1{1-x}}=e^{-1}$.

\pn51. Find $\Lim_{x\to0^+} x^x=
\Lim_{x\to0^+} e^{x\ln x}=e^{\Lim_{x\to0^+} x\ln x}$,
so we compute $\Lim_{x\to0^+} x\ln x=
\Lim_{x\to0^+}\frac{\ln x}{\frac1x}$ and now we may apply 
l'H\^opital's Rule.
$\Lim_{x\to0^+} x\ln x=\Lim_{x\to0^+} 
\frac{\frac 1x}{\frac{-1}{x^2}}=\Lim_{x\to0^+} -x=0$.
Hence $\Lim_{x\to0^+} x^x=e^0=1$.

\pn53. Find
$\Lim_{x\to\infty}\frac{\sqrt{9x+1}}{\sqrt{x+1}}$.
l'H\^opital's Rule applies and $\dd{\sqrt{9x+1}}x=
\frac{9}{2\sqrt{9x+1}}$ and $\dd{\sqrt{x+1}}x=
\frac{1}{2\sqrt{x+1}}$ so we see
$\Lim_{x\to\infty}\frac{\sqrt{9x+1}}{\sqrt{x+1}}=
\Lim_{x\to\infty}\frac{\frac{9}{2\sqrt{9x+1}}}{\frac{1}{2\sqrt{x+1}}}=
9\Lim_{x\to\infty}\frac{\sqrt{x+1}}{\sqrt{9x+1}}=
\frac 9{\Lim_{x\to\infty}\frac{\sqrt{9x+1}}{\sqrt{x+1}}}$.
If $L=\Lim_{x\to\infty}\frac{\sqrt{9x+1}}{\sqrt{x+1}}$ this shows $L^2=9$.
Somewhat more easily,
$L^2=\br3{\Lim_{x\to\infty}\frac{\sqrt{9x+1}}{\sqrt{x+1}}}^2=
\Lim_{x\to\infty}\frac{{9x+1}}{{x+1}}=9$.
Since $\frac{\sqrt{9x+1}}{\sqrt{x+1}}>0$ for large $x$, we see
$L=3$ or the limit does not exist, but we have no theorems to
help us understand which of these is the case.
To actually compute the limit, proceed as follows.
Since $\frac{\sqrt{9x+1}}{\sqrt{x+1}}>0$ for large $x$
$\Lim_{x\to\infty}\frac{\sqrt{9x+1}}{\sqrt{x+1}}=
\sqrt{\Lim_{x\to\infty}\frac{9x+1}{x+1}}\sqrt9=3$.

\pn57. Calculation b) is correct and follows from the usual
quotient rule for limits.
Calculation a) is an incorrect application of l'H\^opital's Rule.

\pn63. Find
$\Lim_{k\to\infty}\br3{1+\frac rk}^k=
\Lim_{k\to\infty}e^{k\ln(1+\frac rk)}=
e^{\lim_{k\to\infty}k\ln(1+\frac rk)}$.
Compute 
$\Lim_{k\to\infty}k\ln(1+\frac rk)=
\lim_{k\to\infty}\frac{\ln(1+\frac rk)}{\frac1k}=
\lim_{k\to\infty}\frac{\frac{\frac{-r}{k^2}}{1+\frac rk}}{\frac{-1}{k^2}}=
r\lim_{k\to\infty}\frac{1}{1+\frac rk}=r$.
Hence 
$\Lim_{k\to\infty}\br3{1+\frac rk}^k=e^r$.

\end
6.6  p. 469:   7, 13, 15, 19, 27, 31, 37, 42, 43, 51, 53, 57, 63
