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\centerline{Mathematics 136 -- Advanced Placement Calculus}
\centerline{Solutions to Ungraded Problems on Problem Set 2}
\centerline{September 22, 2009}
\vskip 10pt
\noindent
2.3/5.  We compute:
\begin{align*}
\lim_{x\to 8} (1 + x^{1/3})(2 - 6x^2 + x^3) 
&= (\lim_{x\to 8} 1 + x^{1/3}) \cdot (\lim_{x\to 8} 2 - 6x^2 + x^3) \text{ by Limit Law 4 }\\
&= (1 + \lim_{x\to 8} x^{1/3}) \cdot (2 - 6 \lim_{x\to 8} x^2 + \lim_{x\to 8} x^3) \text{ by Limit Laws 1, 3}\\
&= (1 + 2)(2 - 6\cdot 64 + 512) \text{ by Limit Law 6}\\
&= 3\cdot 130 = 390.
\end{align*}
\vskip 10pt 
\noindent
2.3/42.  We have by Limit Law 11:
$$
\lim_{v\to c^-} \sqrt{1 - v^2/c^2} = \sqrt{\lim_{v\to c^-}(1 - v^2/c^2)} = 0.
$$
This says that the length of the object appears (to any observer) 
to go to zero as its speed relative to that observer approaches the 
speed of light.  There are two ways to answer the last part:  Physically, 
no object can move faster than $c$, so we must take the limit as 
$v \to c^-$.  Mathematically, the domain of the $L$ function only consists
of $v$ with $-c \le v \le c$.  So again, the limit must be taken as 
$v \to c^-$ (or else we get sqare roots of negative numbers).  
\vskip 10pt
\noindent
2.4/6.  There are infinitely many correct graphs here.  Any graph 
showing $\lim_{x\to -1^-} g(x) = g(-1)$ and $\lim_{x\to 4^+} g(x) = g(4)$
is OK.  For instance we might have jump discontinuities at $x = -1,4$
and the value of $g$ at $-1$ matches the limit from the left, while
the value of $g$ at $4$ matches the limit from the right.
\vskip 10pt
\noindent
2.5/5.  Again, infinitely many correct graphs.  There should be 
a vertical asymptote at $x = 0$, where $f(x) \to -\infty$ from 
both sides (notice the form of the limit as $x \to 0$).  There
should also be horizontal asymptotes at $y = 5$ to the left and
$y = -5$ to the right.
\vskip 10pt
\noindent
2.5/26.  We have 
\begin{align*}
\lim_{x\to\infty} \frac{x+2}{\sqrt{9x^2+1}}
&= \lim_{x\to\infty} \frac{x+2}{\sqrt{9x^2+1}}\cdot  \frac{1/x}{1/x}\\
&= \lim_{x\to\infty} \frac{1+2/x}{\sqrt{9+1/x^2}}\\
&= \frac{1}{3},
\end{align*}
since $2/x \to 0$ and $1/x^2 \to 0$ as $x \to \infty$.  
\vskip 10pt
\noindent
2.6/17.  Of these numbers only $g'(0)$ is negative (the slope
of the tangent at $0$), $g'(-2)$ looks larger (tangent is
steeper) than $g'(2)$ or $g'(4)$.  Hence the order is
$$
g'(0) < 0 < g'(4) < g'(2) < g'(-2).
$$
\noindent
2.7/9.  Since the graph of $f$ is made up of 
straight line segments, the graph $y = f'(x)$ 
will be ``piecewise constant.''  Starting from
the left, $y = f'(x)$ should be a negative
constant $c_1$ up to $x = -a$ (say), a positive constant
$c_2$ from $x = -a$ to $x = 0$, the negative constant 
$-c_2$ from $x = 0$ to $x = a$, and $-c_1$ from $x = a$ on.
The derivative graph has jump discontinuities at 
$x = -a, 0, a$, and $f'(x)$ is undefined there
(so open circles on both sides on $y = f'(x)$).   
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