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\centerline{Mathematics 242 -- Principles of Analysis}
\centerline{Exam 3 -- Solutions}
\centerline{December 6, 2004}
\bigskip
\noindent
I. A)  Let $f : D \to {\bf R}$ and let $c$ be an accumulation
point of $D$.  We say $\lim_{x\to c} f(x) = L$ if
for all $\varepsilon > 0$, there exists $\delta > 0$ 
such that $|f(x) - L| < \varepsilon$ for all $x \in D$
with $0 < |x - c| < \delta$.

B)  The domain of $f$ is $D = {\bf R}$ in this case.
Given $\varepsilon > 0$, let $\delta = \sqrt{\varepsilon}$.
Then for all $x$ with $0 < |x - 2| < \sqrt{\varepsilon}$, 
$$|f(x) - (-2)| = |x^2 - 4x + 4| = |x - 2|^2 < (\sqrt{\varepsilon})^2
= \varepsilon.$$
Hence the definition is satisfied and $\lim_{x\to 2} x^2 - 4x + 2 = -2$.

C)  The limit $\lim_{x\to 0} {x^2 \over \sin(2x^2)}$
is indeterminate of the form $0/0$ and the other hypotheses
for L'Hopital's Rule are satisfied, so we apply L'Hopital 
to see:
$$\eqalign{
\lim_{x\to 0} {x^2 \over \sin(2x^2)}
&= \lim_{x\to 0} {2x \over 4x\cos(2x^2)}\cr
&= \lim_{x\to 0} {1\over 2\cos(2x^2)}\cr
&= 1/2\cr}$$
(Note: The limit on the right side of the first line
here is also indeterminate $0/0$.  So it could be
evaluated by applying L'Hopital a second time.  But
canceling the $x$'s is certainly easier!)
\bigskip
\noindent
II.  A)  Intermediate Value Theorem.  Let $f$ be
continuous on $[a,b]$.  (Note: no differentiability 
hypotheses are necessary here!)  Assume $f(a) \ne f(b)$.
Then for all $y$ between $f(a)$ and $f(b)$, there exists
at least one $x \in [a,b]$ such that $f(x) = y$.

B)  $f(x) = a_1 + a_2 \sin(x)$ is continuous for all
$x\in {\bf R}$ (since $f$ is differentiable, for instance!)
So the IVT applies on the interval $[-\pi/2,\pi/2]$.  We have
$f(-\pi/2) = a_1 + a_2 \sin(-\pi/2) = a_1 - a_2 < 0$ since
$a_2 > a_1$.  On the other hand $f(\pi/2) = a_1 + a_2 > 0$
since $a_1, a_2 > 0$.  Hence $f$ changes sign on this
interval and by the IVT, there exists $c \in [-\pi/2,\pi/2]$
such that $f(c) = 0$. 

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\noindent
III.  Here's the proof we did in class. (This is slightly different
from the one in the text;  Lay doesn't assume
the partition is regular, but argues that if 
{\it all} the smaller intervals in the partition
are sufficiently small, then the difference
between the upper and lower sums
can be made arbitrarily small.  Both proofs are correct of course!).  
Let $P_n = \{a=x_0,x_1,\ldots,x_{n-1},x_n=b\}$ 
be the regular partition of $[a,b]$ with 
$\Delta x = (b-a)/n$.  Then since $f$ is monotone
increasing,
$$U(f,P_n) = \sum_{i=1}^n M_i \Delta x = \sum_{i=1}^n f(x_i) \Delta x$$
and 
$$L(f,P_n) = \sum_{i=1}^n m_i \Delta x = \sum_{i=1}^n f(x_{i-1}) \Delta x$$
In the difference, $U(f,P_n) - L(f,P_n)$, the sum ``telescopes''
and we get
$$U(f,P_n) - L(f,P_n) = (f(b) - f(a))\Delta x = (f(b) - f(a))(b - a)/n$$
Given any $\varepsilon > 0$, take $n > (f(b) - f(a))(b - a)/\varepsilon$.
Then $U(f,P_n) - L(f,P_n) < \varepsilon$.  (This shows that $f$
is integrable by the result of Exercise 29.9 on the last problem 
set -- the idea is that since $U(f,P_n)$ and $L(f,P_n)$ 
can be made arbitrarily close by taking $n$ large enough,
$U(f) = \inf_P \{U(f,P)\}$ and $L(f) = \sup_P \{L(f,P)\}$
must be equal.

\bigskip
\noindent
IV.  A)  This is FALSE.  $f$ is not continuous at $0$
(because $\lim_{x\to 0^-} f(x) = -2$, while
$\lim_{x\to 0^+} f(x) = +2$).  So
$f$ is not differentiable at $0$ either.  This can also
be seen by trying to compute the limit of the difference
quotient for $f$.  If $x \ge 0$, $f(x) = 2 - x - x^3$,
so 
$$\lim_{x\to 0^+} {f(x) - f(0)\over x - 0} = 
\lim_{x\to 0^+} {-x - x^2 \over x} = \lim_{x\to 0^+} -1 - x = -1.$$
But 
$$\lim_{x\to 0^-} {f(x) - f(0)\over x - 0} = 
\lim_{x\to 0^-} {-x - 4 \over x} = \lim_{x\to 0^-} -1 - {4\over x},$$
which does not exist (we might say the limit goes to $+\infty$).
Hence $f'(0)$ does not exist.
\bigskip
\noindent
B)  This is TRUE. Full credit was given for the following 
argument:  Split the integral at $x = 0$.  Since $f$
is continuous and monotone decreasing on $[-2,0]$ and 
$[0,2]$, then
we know $f$ is integrable on each half by the monotone decreasing
case of the theorem from question III.  Then 
$$\int_{-2}^2 f(x)\ dx = \int_{-2}^0 f(x)\ dx + \int_0^2 f(x)\ dx$$
exists too.  (You could also get the same result by considering
partitions containing $x_k = 0$ for some $k$, so the
upper and lower sums for $f$ on the whole interval
split into upper and lower sums for 
$f$ on each half.  Since $f$ is integrable on each half
of the interval, the upper and lower sums can be 
made arbitrarily close.)  This reasoning suggests an important
general statement:  {\it If $f$ is continuous except at  
a finite number of points in $[a,b]$, or if $[a,b]$ 
can be split into a finite number of intervals 
on which $f$ is monotonic, then $f$ is 
integrable on $[a,b]$.}  This can be proved in the same
way -- split into subintervals, and apply the
appropriate theorem 
on each piece.

\bigskip
\noindent
C) This is TRUE.  The {\it conclusion} of the MVT is that there exists
some $c$ in the interval $[a,b]$ such that
$f'(c) = {f(b) - f(a) \over b - a}$.  The hypotheses
of the MVT are not satisfied for this function
on the interval $[-2,2]$ (see part A).
HOWEVER, note that 
${f(2) - f(-2) \over 2 - (-2)} = {-8 - 0\over 4} = -2$.
Moreover, $f$ is differentiable for all $x \ne 0$, 
and 
$$f'(x) = \cases{-1 & if $x < 0$\cr
                 -1 - 3x^2 & if $x > 0$\cr}$$
We have $-2 = -1 - 3x^2$ if $3x^2 = 1$, or $x = \pm 1/\sqrt{3}$.
$x = 1/\sqrt{3}$ is in our interval, so the conclusion of 
of the MVT is satisfied!

\bigskip
\noindent
D)  This is FALSE.  The {\it conclusion} of the
EVT is that, writing 
$M = \sup\{f(x) : x \in [a,b]\}$ and
$m = \inf\{f(x) : x \in [a,b]\}$,
there exist $c_1, c_2 \in [a,b]$ such that
$f(c_1) = M$ and $f(c_2) = m$.  Again, the 
hypotheses of the EVT are not satisfied on 
$[-2,1]$ (because $f$ is not continous at $0$).
Here, by considering the graph of $f$, we 
can see $M = 2 = f(0)$, so $f$ does attain
its maximum on the interval.  But, $m = -2$
since $\lim_{x\to 0^-} f(x) = -2$.
But there is no $x \in [-2,1]$ such that
$f(x) = -2$ (note $f(0) = +2$, not $-2$ 
by the definition of $f(x)$, and the smallest
value of $f$ for $x$ between $0$ and $1$ is 
$f(1) = 0$.)

\bigskip
\noindent
General comment on parts C, D:  The point of these 
questions was not only to test your knowledge of the
relevant theorems, but also to test your understanding
of the underlying logic of mathematical statements.
Even if the hypotheses of a theorem are not satisfied,
it is possible for the conclusion to be true (see part C).
However, there is no guarantee that the conclusion must
be true (see part D).  

 
\bye


