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\centerline{Mathematics 242 -- Principles of Analysis}
\centerline{Problem Set 6 -- {\bf due:} Monday, March 24}
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{\it 'A' Section}
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1.  Determine whether each of the following limits exists using the ``big theorem''
for function limits and other results from section 3.2 of the text as needed.  
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\item{(a)}  $\lim_{x \to 1} x^2 - 5x + 3$
\item{(b)}  $\lim_{x \to {1\over 3}} x + {1\over x^2}$
\item{(c)}  $\lim_{x \to 1} {x^3 - 1\over x^2 - 1}$
\item{(d)}  Let 
$$f(x) = \cases{x^{1/3} \sin\left({1\over x}\right) & if $x \ne 0$\cr
                          3 & if $x = 0$\cr}$$
and consider $\lim_{x \to 0} f(x)$.   
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2.  Which of the functions in question 1 are {\it continuous}  at the 
indicated $c$ in the limits there?  Explain.
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3.  True-False.  For the true statements, give a short proof.  For
the false statements give a counterexample.
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\item{(a)}  If $\lim_{x \to 1} f(x) = e - {28\over 10}$, then there exists
a $\delta > 0$ such that $f(x) < 0$ for all $x$ with $0 < |x - 1| < \delta$.  
\item{(b)}  If $|f(x)| \le x^3$ for all $x$ and $\lim_{x\to 2} f(x)$ exists, 
then $\lim_{x\to 2} f(x) \le 8$.  
\item{(c)}  Let $f : {\bf R} \to {\bf R}$ be defined by this rule:
$$f(x) = \cases{2x & if $x$ is rational\cr
                -2x & if $x$ is irrational.\cr}$$
Then $\lim_{x \to 0} f(x)$ exists and equals $0$.  
\item{(d)}  If $f(x) < g(x)$ on a deleted neighborhood of $c$, 
$\lim_{x \to c} f(x) = L$, and $\lim_{x\to c} g(x) = M$, then $L < M$.  
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{\it 'B' Section}
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1.  Show that your answers for parts a and d of 1 on the A section are 
correct using the $\varepsilon,\delta$ definition ({\it not} the ``big theorem'' or
other results from Chapter 3, section 1 of the text.)
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2.  Assume that $\lim_{x\to c} f(x) = L$.  
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\item{(a)}  Show that there exists a constant $B$ and $\delta > 0$ such 
that $|f(x)| \le B$ for all $x$ in the deleted neighborhood 
$\{x\in {\bf R} \mid 0 < |x - c| < \delta\}$.
\item{(b)}  Using part (a), {\it not} the limit product rule,  
show that $\lim_{x\to c} (f(x))^n = L^n$ for all integers $n \ge 1$.  
\item{(c)}  Assume that $f(x) \ge 0$ on some deleted neighborhood of $x = c$.
Show that $$\lim_{x\to c} \sqrt{f(x)} = \sqrt{L}.$$  
({\it Hint:}  It may help to treat the cases $L = 0$ and $L \ne 0$ separately.)  
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3.  In this problem you will show that 
$$\lim_{\theta\to 0} {\sin(\theta)\over \theta} = 1.$$  
For $0 < \theta < {\pi \over 2}$, the point $P = (\cos(\theta),\sin(\theta)) = (x,y)$ lies on the 
arc of the unit circle $x^2 + y^2 = 1$ in the first quadrant.  
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\item{(a)}  Let $O = (0,0)$, $Q = (\cos(\theta),0)$, and $R = (1,0)$.  (Draw a picture!)
By considering the areas of the triangle $\Delta OQP$
and the circular sector $ORP$, deduce that if $0 < \theta < {\pi \over 2}$, then
$\sin(\theta)\cos(\theta) \le \theta$.   (You may use ``intuitively reasonable'' facts about areas
such as the statement that if one plane region ${\cal R}$ is completely completely
contained in a second region ${\cal S}$, then ${\rm area}({\cal R}) \le {\rm area}({\cal S})$.)  
\item{(b)}  Now take the tangent line to the circle at $R$ (a vertical line), and 
let $S = (1, \tan(\theta))$ be the intersection of that line and the radius $OP$ (extended).
Considering the areas of the triangle $\Delta ORS$ and the sector $ORP$ as above, 
explain why $\theta \le \tan(\theta)$.
\item{(c)}  Combine parts (a) and (b) to deduce that if $0 < \theta < {\pi \over 2}$, then 
$$\cos(\theta) \le {\sin(\theta) \over \theta} \le {1\over \cos(\theta)}.$$
\item{(d)}  Using the one-sided form of Theorem 3.2.9 (The Limit Squeeze Theorem), show that
$$\lim_{\theta \to 0^+} {\sin(\theta)\over \theta} = 1.$$
(You will need to use the fact that $\cos(\theta)$ is continuous at $\theta = 0$.)
\item{(e)}  Now, for $-{\pi\over 2} < \theta < 0$, show that 
${\sin(\theta)\over \theta} = {\sin(|{\theta}|)\over |\theta|}$ and use this
to see that 
$$\lim_{\theta \to 0^-} {\sin(\theta)\over \theta} = 1$$ 
as well.  
\item{(f)}  Finally, explain how parts (d) and (e) combine to show the statement at the 
start of the problem.
\bye
