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\centerline{Mathematics 242 -- Principles of Analysis}
\centerline{Exam 2 -- Solutions}
\bigskip
\noindent
I. 
\item{A)}  See the class notes or the text.
\item{B)}  We want to show that $\lim_{x\to 2} x^2 - 5x = -6$.
First note that 
$$x^2 - 5x - (-6) = (x - 2)(x - 3),\ {\rm so}\ |x^2 - 5x + 6| = |x - 2||x - 3|$$
If $0 < |x - 2| < \delta$, and $\delta < 1$, then $1 < x < 3$, so 
$-2 < x - 3 < 0$.  Hence $|x - 3| < 2$.  Therefore, if $\epsilon > 0$
is any given positive number, $\delta < {\rm min}(1,\epsilon/2)$,
and $0 < |x - 2| < \delta$, then we have
$$\eqalign{|x^2 - 5x + 6| &= |x - 2||x - 3|\cr
                          &< (\epsilon/2)\cdot 2\cr
                          &=\epsilon\cr}$$
\bigskip
\noindent
II.  See the class notes (or the notes from the review session, or the 
text).
\bigskip
\noindent
III.  
\item{A)} The answer is {\it no}.   We have
$$f'(0) = \lim_{x \to 0} {f(x) - f(0)\over x - 0} = 
\lim_{x\to 0}{f(x) - 1\over x},$$
provided that the limit exists.  But if we let $x \to 0^{-}$, then 
since $f(x) = 1$ for all $x < 0$, we see
$$\lim_{x\to 0^-} {f(x) - 1\over x} - \lim_{x\to 0^-} {1 - 1\over x} = 0.$$
On the other hand letting $x \to 0^+$, we have
$$\lim_{x\to 0^+} {f(x) - 1\over x} = \lim_{x\to 0^+} {1 + x + x^2 - 1\over x}
= \lim_{x\to 0^-} 1 + x = 1$$
Since the two one-sided limits are not the same, $f'(0)$ does not 
exist.  
\item{B)}  Full credit was given for the answer ``$f$ is integrable, 
since it is continuous on $[-1,1]$.''  This can also be proved directly
without invoking that theorem, by considering partitions as indicated
in the problem statement.  Note that $f$ is {\it strictly increasing --
not constant} on the interval from $0$ to $1$, so we will {\it not be 
able to get the difference between an upper and lower sum to be arbitrarily
small by picking partitions with only one or two subintervals}.  Instead, we
note that if $P_n = \{-1,x_1=0,x_2,\ldots,x_n = 1\}$ where 
$P' = \{0,x_2,\ldots, 1\}$
is a regular partition of $[0,1]$ (with $n-1$ smaller
intervals), then since $f(x)$ is strictly increasing
on $[0,1]$ we will have:
$$\eqalign{
U_{P_n}(f) &= 1\cdot 1 + \sum_{i=2}^n f(x_i)\Delta x\cr
L_{P_n}(f) &= 1\cdot 1 + \sum_{i=2}^n f(x_{i-1})\Delta x,\ {\rm so}\cr
U_{P_n}(f) - L_{P_n}(f) &= (f(1) - f(0))\Delta x\cr
                &= 2\Delta x\cr}$$
This can be made arbitrarily small by picking $\Delta x$ small enough
(that is taking $n = {1\over \Delta x} + 1$ large enough).  Hence
$f$ is integrable on $[-1,1]$.
\bigskip
\item{C)}  The value of the integral $\int_{-1}^1 f(x)\ dx$ may be
found without the Fundamental Theorem of Calculus by taking the limit
of the upper sums for a partition as in part B.  For simplicity
we write $m = n-1$ (the number of smaller intervals after the first, 
``big'' one from $-1$ to $0$):
$$\eqalign{
\int_{-1}^1 f(x)\ dx &= \lim_{n\to\infty} U_{P_n}(f)\cr
&=\lim_{m\to\infty} 1 + \sum_{i=1}^m \left(1 + {i\over m} + \left({i\over m}\right)^2\right){1\over m}\cr
&=1 + \lim_{m\to\infty} \left( {1\over m} \sum_{i=1}^m 1 + {1\over m^2} \sum_{i=1}^m i
+ {1\over m^3} \sum_{i=1}^m i^2\right)\cr
&= 1 +\lim_{m\to\infty} \left(1 + {1\over m^2}{m(m+1)\over 2}
+ {1\over m^3} {m(m+1)(2m+1)\over 6}\right)\cr
&=1 + 1 + {1\over 2} + {1\over 3}\right)\cr
&= {17\over 6}\cr}$$
\bigskip
\noindent
IV.  True-False.
\bigskip
\item{A)}  This statement is FALSE:
$$\cases{\lim_{x\to 0^-} {\sin(x)\over |x|} = -1& but\cr
          \lim_{x\to 0^+} {\sin(x)\over |x|} = +1&\cr}$$
Therefore, ${\sin(x)\over |x|}$ has a jump discontinuity at $x = 0$,
not a removable discontinuity.
\bigskip
\item{B)}  This statement is also FALSE.  The IVT has no converse statement.
For example the function:
$$f(x) = \cases{x^2 &if $0 \le x \le 1$\cr
                {1\over 2} & if $1 < x \le 2$}$$
has the intermediate value property on $[0,2]$ -- for every $y$ between
$y = m = 0$ and $y = M = 1$, there is some $x$ such that $f(x) = y$.
But $f$ is not continuous on $[0,2]$ because of the jump at $1$.
\bigskip
\item{C)}  This statement is TRUE.  By hypothesis, $f''(x)$ exists
for all $x$.  This implies that $f'(x)$ exists and is continuous for
all $x$.  Similarly, $f(x)$ exists and is continuous for all $x$.
We may assume $x_1 < x_2 < x_3$.  By Rolle's Theorem (MVT) on 
$[x_1,x_2]$ there is some $c_1 \in (x_1,x_2)$ such that $f'(c_1) = 0$.
Similarly, on $[x_2,x_3]$, there is some $c_2$ such that $f'(c_2) = 0$.
But now by Rolle's Theorem applied to $f'(x)$ on the interval
$[c_1,c_2]$, there is some
$d \in (c_2,c_2)$ such that $(f')'(d) = f''(d) = 0$.
\bye