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\centerline{Mathematics 242 -- Principles of Analysis}
\centerline{Exam 3 Solutions -- December 2, 2005}
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\noindent
I.  $f(x) = x^2 - 6x + 3$. 
\item{A)}  Let $\varepsilon > 0$ and $\delta < \min \{1, \varepsilon/5 \}$.  
If $|x - 1| < \delta < 1$, then $0 < x < 2$, so $|x - 5| < 5$.  Then
since we also have $|x - 1| < \varepsilon/5$:
$$|f(x) - (-2)| = |x^2 - 6x + 5| = |x - 1||x - 5| < (\varepsilon/5)\cdot 5 = \varepsilon.$$
This shows $\lim_{x\to 1} f(x) = -2$.
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\item{B)}  Note that $f'(x) = 2x - 6 \ge 0$ on the interval $[3,4]$.  Therefore
$f$ is monotone increasing.  By the result from question C on Discussion 3, 
$f$ is integrable on that interval.  (Alternate way:  $f$ is continuous
on ${\bf R}$, so by the theorem we proved in class on 11/30, $f$ is integrable
on every finite interval.)
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\noindent
II.  
\item{A)}  IVT:  Let $f$ be continuous on $[a,b]$.  If $L$ is any real
number (strictly) between $f(a)$ and $f(b)$, then there exists 
$c\in (a,b)$ such that $f(c) = L$.
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\item{B)}  $f(x)$ is continuous because $f'(x) = -a_2 \sin(x)$ exists
for all real $x$.  Moreover $f(0) = a_1 + a_2 > 0$ while $f(\pi) = a_1 - a_2 < 0$.
Taking $L = 0$ in the IVT, there exists $c \in (0,\pi)$ such
that $f(c) = 0$.
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\noindent
III.  Since $s = \sup f([a,b])$, if $\varepsilon > 0$, then 
$s - \varepsilon$ is not an upper bound for $f([a,b])$.  So 
there exists $x \in [a,b]$ such that $s - \varepsilon < f(x) \le s$.
Apply this with $\varepsilon = {1\over n}$ for each $n \in {\bf N}$.
This gives a sequence $(x_n)$ in $[a,b]$ such that 
$s - {1\over n} < f(x_n) < s$ for all $n\in {\bf N}$.
Since $[a,b]$ is a bounded interval, the Bolzano-Weierstrass
theorem implies that there is a subsequence $x_{n_k}$
of this sequence converging to some $c$.  By the above
$\lim_{k\to\infty} f(x_{n_k}) = s$.
Since the 
interval is closed, the limit $c$ is also in the interval
$[a,b]$.  But then by the sequential version of continuity
$$s = \lim_{k\to\infty} f(x_{n_k}) = f(c).$$
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\noindent
IV.  
\item{A)}  This is FALSE.  Let $x_n$ be any sequence of
rational numbers converging to 3.  Then $\lim_{n\to\infty} f(x) = 9$.
On the other hand, if $y_n$ is a sequence of irrational numbers
converging to 3, then $\lim_{n\to\infty} f(y_n) = 0$.  Since
these two sequential limits are different the limit of the
function does not exist.
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\item{B)}  This is TRUE.  We have
$${f(x) - f(0) \over x - 0} = \cases{{x^2\over x} = x & if $x$ is rational\cr
                                        0 & if $x$ is irrational\cr}$$
Either way, if $|x| < \delta = \varepsilon$, then 
$$\left|{f(x) - f(0) \over x - 0} - 0\right| < \varepsilon$$
whenever $|x| < \varepsilon$.  So the limit of the difference 
quotient exists and 
$$f'(0) = \lim_{x\to 0} {f(x) - f(0) \over x - 0} = 0.$$
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\item{C)}  This is FALSE.  If we pick any partition $P = \{x_i : i = 0,\ldots, n\}$ of the 
interval, then each subinterval contains rational numbers 
arbitrarily close to the right-hand endpoint of the subinterval.
Hence $M_i = x_i^2$.  On the other hand $m_i = 0$ since
every subinterval also contains irrational numbers.
Since all the $x_i^2 > 1$, The difference 
$$U(f,P) - L(f,P) = \sum_{i=1}^n (M_i - m_i) \Delta x_i = \sum_{i=1}^n x_i^2 \Delta x_i
> \sum_{i=1}^n \Delta x_i = 1.$$
This shows $f$ is not integrable on $[1,2]$.
\item{D)}  This is FALSE.  $s = \sup f([-\sqrt{2},\sqrt{2}]) = 2$
but there is no $x$ in the interval where $f(x) = 2$.  (Note $\pm\sqrt{2}$ 
are irrational,
so $f(\pm\sqrt{2}) = 0$).

\bye