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\begin{center}
\textbf{Holy Cross College, Fall Semester, 2003} \\
\textbf{Math 131, Professor Hwang} \\
L'H\^{o}pital's Rule Worksheet Answers
\end{center}

\noindent Calculate the following limits:
  \[
  \lim_{x\to1}\frac{\ln x}{x-1}=1,\qquad
  \lim_{x\to0}\frac{1-\cos x}{x}=0,\qquad
  \lim_{x\to0}\frac{e^x-1-x}{x^2}=\frac{1}{2},\qquad
  \lim_{x\to0}\frac{x-\sin x}{x^3}=\frac{1}{6}.
  \]

  \vfil

  \[
  \lim_{x\to\infty}\frac{x}{e^x},\qquad
  \lim_{x\to\infty}\frac{x^2}{e^x},\qquad
  \lim_{x\to\infty}\frac{x^n}{e^x} \quad(n>0).
  \]
  Each limit is zero. The reason is that each application of
  l'H\^{o}pital's rule decreases the exponent in the numerator by~1,
  while the denominator remains the same; eventually the exponent
  becomes zero or negative.

  \vfil

  \[
  \lim_{x\to\infty}\frac{\ln x}{x},\qquad
  \lim_{x\to\infty}\frac{\ln x}{\sqrt{x}},\qquad
  \lim_{x\to\infty}\frac{\ln x}{x^{0.001}}.
  \]
  Each limit is zero. In general, if $p>0$, then
  \[
  \lim_{x\to\infty}\frac{\ln x}{x^p}
  =\lim_{x\to\infty}\frac{1/x}{px^{p-1}}
  =\lim_{x\to\infty}\frac{1}{px^p}=0.
  \]

  \vfil

\noindent We say that ``$g$~grows faster than~$f$'' if
$\lim\limits_{x\to\infty}\frac{f(x)}{g(x)} = 0$. Put the functions
below in increasing order of growth rate.

\[
e^x,\quad \ln(\ln x),\quad x^{100},\quad e^{x^2}, \quad x^{1/2},\quad
\ln x,\quad e^{2x},\quad 1000x^{10},\quad x^x
\]

  The previous sets of questions underscore that every exponential
  function grows faster than every power function, and that every
  power function with positive exponent grows faster than the
  natural logarithm. Writing $f\ll g$ to denote the fact that
  $g$~grows faster than~$f$ (or ``$g$ dominates~$f$''), we have
\[
\ln(\ln x)\ll 
\ln x\ll 
x^{1/2}\ll
1000x^{10}\ll 
x^{100}\ll 
e^x\ll 
e^{2x}\ll 
x^x \ll
e^{x^2}
\]
The power functions and exponentials should be fairly easy to
group. The trickiest cases are $\ln(\ln x)=:\ln u$, which grows more
slowly than~$\ln x=:u$, and
\[
x^x=e^{x\ln x},
\]
which grows more rapidly than $e^{2x}$, but more slowly than~$e^{x^2}$.

\eject

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