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\centerline{Mathematics 242 -- Principles of Analysis}
\centerline{Solutions for Problem Set 6 -- {\bf due:} Monday, March 24}
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\noindent
{\it 'A' Section}
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\noindent
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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\noindent

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\item{(a)}  $\lim_{x \to 1} x^2 - 5x + 3$
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\item{} {\it Solution:}  By parts 1 and 2 of the ``big theorem,'' the limit is 
$$\lim_{x\to 1} x^2 - 5 \lim_{x\to 1} x + 3 = 1 - 5 + 1 = -1.$$   
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\item{(b)}  $\lim_{x \to {1\over 3}} x + {1\over x^2}$
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\item{} {\it Solution:}  By parts 1 and 3 of the ``big theorem,''
the limit is ${1/3} + 9 = {28\over 3}$. 
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\item{(c)}  $\lim_{x \to 1} {x^3 - 1\over x^2 - 1}$
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\item{} {\it Solution:}  The limit is ${3\over 2}$.  Proof:  For $x \ne 1$, we see
$${x^3 - 1\over x^2 - 1} = {(x^2 + x + 1)(x - 1)\over (x - 1)(x + 1)} = {x^2 + x + 1\over x + 1}$$
This shows $\lim_{x \to 1} {x^3 - 1\over x^2 - 1} = \lim_{x\to 1} {x^2+x+1\over x+1} = {3\over 2}$.  
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\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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\item{} {\it Solution:}  The limit is $0$ by the limit squeeze theorem.
We have $-1\le \sin\left({1\over x}\right) \le 1$ for all $x \ne 0$.  So 
$$-x^{1/3} \le f(x) \le x^{1/3}$$
for all $x \ne 0$.  Since $\lim_{x \to 0} -x^{1/3} = \lim_{x\to 0} x^{1/3} = 0$,
$\lim_{x \to 0} f(x) = 0$ also.       
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\noindent
2.  Which of the functions in question 1 are {\it continuous}  at the 
indicated $c$ in the limits there?  Explain.
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\noindent
{\it Solution:}  The functions in parts (a) and (b) of problem 1 are continuous at the 
given $c$ since $\lim_{x \to c} f(x) = f(c)$.  The function in part (c) is not 
continuous at $2$ since $f(1)$ is not defined.  The function in part (d) is not continuous
at $x = 0$ since $\lim_{x \to 0} f(x) = 0 \ne 3 = f(0)$ (by the definition of $f(x)$).  
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\noindent
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 $0 < |x - 1| < \delta$.  
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\item{} This is TRUE.  The reason is that $a = e - {28\over 10} < 0$ (since 
$e \doteq 2.71828 < 2.8$).  So if we 
take $\varepsilon = |a|/2$, then there is a corresponding 
$\delta > 0$ such that for $x$ with $0 < |x - 1| < \delta$, 
$$\left|f(x) - a\right| < |a|/2.$$
But this implies $f(x) < a + |a|/2 = a/2 < 0$ for all such $x$.   
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\item{(b)}  If $|f(x)| \le x^3$ for all $x$ and $\lim_{x\to 2} f(x)$
exists, then $\lim_{x\to 2} f(x) \le 8$.  
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\item{} {\it Solution:}  This is TRUE.  We have $f(x) \le |f(x)| \le x^3$
so $\lim_{x\to 2} f(x) \le \lim_{x\to 2} x^3 = 8$, using Theorem 3.2.8.
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\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 $0$.
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\item{} {\it Solution:}  This is TRUE.  Given any $\varepsilon > 0$, let $\delta = \varepsilon/2$.  
Then for all $x$ with $0 < |x - 0| < \delta = \varepsilon/2$, we have $|2x| = |-2x| = 2|x|$ for 
rational and irrational $x$, so 
$$|f(x)| = 2|x| < 2 {\varepsilon\over 2} = \varepsilon.$$
This shows the limit is $0$ as claimed.
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\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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\item{} {\it Solution:}  This is FALSE.  Counterexample:  Let $f(x) = x^4$ and $g(x) = x^2$.
Then $f(x) < g(x)$ for all $x$ with $0 < |x - 0| < 1$.  But $\lim_{x \to 0} f(x) = 0 = \lim_{x \to 0} g(x)$. 
(The statement would be true if it said $L \le M$.)
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\noindent
{\it 'B' Section}
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\noindent
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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\noindent
{\it Solution:}  Part (a) first.   Given $\varepsilon > 0$,
let $\delta = \min(1, \varepsilon/4)$.  For all $x$ with $0 < |x - 1| < \delta < 1$, we have
$0 < x < 2$, so $|x - 4| < 4$ and hence
$$\eqalign{
|x^2 - 5x + 3 - (-1)| &= |x^2 - 5x + 4|\cr
&= |x - 4||x - 1|\cr
&< 4 \cdot {\varepsilon \over 4} \cr
&= \varepsilon.\cr}$$
This shows $\lim_{x\to 1} x^2 - 5x + 3 = -1$.
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\noindent
Part (b).  Given $\varepsilon > 0$, let $\delta = \varepsilon^3 > 0$.  Then 
For all $x$ with $0 < |x - 0| < \delta = \varepsilon^3$, we have 
$$|f(x) - 0| = \left|x^{1/3} \sin\left({1\over x}\right)\right| = |x|^{1/3} \left|\sin\left({1\over x}\right)\right|
\le |x|^{1/3} < (\varepsilon^3)^{1/3} = \varepsilon.$$
This shows $\lim_{x\to 0} f(x) = 0$.  
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\noindent
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 
$\{x\in {\bf R} \mid 0 < |x - c| < \delta\}$.
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\item{} {\it Solution:}  Since $\lim_{x\to c} f(x) = L$, letting $\varepsilon = 1$, there
is a corresponding $\delta > 0$ such that $|f(x) - L| < 1$ for all $x$ in the deleted
neighborhood defined by $0 < |x - c| < \delta$.  But for those $x$, $L - 1 < f(x) < L + 1$,
so $|f(x)| \le {\rm max}(|L+1|, |L-1|)$.  We can take $B = {\rm max}(|L+1|, |L-1|)$.  
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\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 $n \ge 1$.
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\item{} {\it Solution:}  Let $B$ and $\delta_0$ be as in part (a).  That is
assume that $|f(x)| \le B$ for all $x$ with $0 < |x - c| < \delta_0$.  Given
$\varepsilon$, since $\lim_{x \to c} f(x) = L$, we have $|f(x) - L| < \varepsilon/M,$ 
where 
$$M = B^{n-1} + B^{n-2} |L| + \cdots +  B |L|^{n-2} + |L|^{n-1}$$
 for all $x$
with $0 < |x - c| < \delta_1$ for some $\delta_1 > 0$.  
Let $\delta = \min(\delta_0, \delta_1)$.  Then 
for all $x$ with $0 < |x - c| < \delta$, we have (using the triangle inequality on the
second factor on the right side):
$$\eqalign{
|(f(x))^n - L^n| &= |f(x) - L| |(f(x))^{n-1} + (f(x))^{n-2} L + \cdots + f(x) L^{n-2} + L^{n-1}|\cr
&\le |f(x) - L| (|f(x)|^{n-1} + |f(x)|^{n-2} |L| + \cdots + |f(x)| |L|^{n-2} + |L|^{n-1})\cr
&< {\varepsilon\over M} (B^{n-1} + B^{n-2} |L| + \cdots + B |L|^{n-2} + |L|^{n-1})\cr
&= {\varepsilon \over M}\cdot M = \varepsilon.\cr}$$
This shows $\lim_{x\to c} (f(x))^n = L^n$.  
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\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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\item{} {\it Solution:}  First suppose $L = 0$.  Then for all $\varepsilon > 0$, there
exist corresponding $\delta > 0$ such that $|f(x)| < \varepsilon^2$ for all $x$ with $0 < |x - c| < \delta$.
But then for the same $x$, we have $|\sqrt{f(x)}| < \varepsilon$.  So $\lim_{x \to c} \sqrt{f(x)} = 0 = \sqrt{0}$.  
Now assume $L \ne 0$ (so $L > 0$).  Given $\varepsilon > 0$, there exists $\delta > 0$
such that $|f(x) - L| < \varepsilon\sqrt{L}$ for all $x$ with $0 < |x - c| < \delta$.  For these $x$, 
$$\eqalign{|\sqrt{f(x)} - \sqrt{L}| 
&= {|f(x) - L| \over f(x) + \sqrt{L}}\cr
&\le  {|f(x) - L| \over \sqrt{L}}\cr
&< {\varepsilon \sqrt{L} \over \sqrt{L}}\cr
&=\varepsilon.\cr}$$
This shows $\lim_{x \to c} \sqrt{f(x)} = \sqrt{L}$.  
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\noindent
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 ${\rm area}({\cal R}) \le {\rm area}({\cal S})$.)  
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\item{} {\it Solution:}  The area of the triangle $\Delta OQP$ is ${1\over 2} \sin(\theta)\cos(\theta)$.
The area of the sector is ${1\over 2} \theta$, since the area of the circular sector with angle $\Theta$
of a circle of radius $r$ is ${\Theta r^2\over 2}$.  Since the sector completely contains the 
triangle, the desired inequality follows.  
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\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 $\theta \le \tan(\theta)$.
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\item{} {\it Solution:}  The area of the triangle $\Delta ORS$ is ${1\over 2} \tan(\theta)$.  (By
the way, in case you never have seen this before, this is the reason that the tangent function is
known by that name!)  This time the sector $ORP$ is completely contained in the triangle,
so the inequality follows again.  
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\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{} {\it Solution:}  Combining parts (a) and (b), we see
$$\sin(\theta)\cos(\theta) \le \theta \le \tan(\theta) = {\sin(\theta)\over \cos(\theta)}.$$
Since $\sin(\theta) > 0$ for the $\theta$ in this range, it follows that
$$\cos(\theta) \le {\theta\over \sin(\theta)} \le {1\over \cos(\theta)}$$
and the desired inequalities follow by taking reciprocals.  
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\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 $\theta = 0$.)
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\item{} {\it Solution:}  Since $\cos(\theta)$ is continuous at $\theta = 0$, we have 
$\lim_{\theta\to 0^+} \cos(\theta) = 1$ and hence $\lim_{\theta\to 0^+} {1\over \cos(\theta)} = 1$
as well.  By the one-sided version of the Limit Squeeze Theorem, 
$$\lim_{\theta\to 0^+} {\sin(\theta)\over \theta} = 1$$
also.  
\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.  
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\item{} {\it Solution:}  This follows from the fact that $\sin$ is an odd function.  If $\theta < 0$, 
then 
$${\sin(|\theta|)\over |\theta|} = {\sin(-\theta) \over -\theta} = {-\sin(\theta) \over -\theta} = {\sin(\theta)\over \theta}.$$ 
By letting $\varphi = |\theta| > 0$, from part (d) we see that
$$\lim_{\theta\to 0^{-1}} {\sin(\theta)\over \theta} = \lim_{\theta\to 0^-} {\sin(|\theta|) \over |\theta|}
= \lim_{\varphi\to 0^+} {\sin(\varphi)\over \varphi} = 1.$$ 
\item{(f)}  Finally, explain how parts (d) and (e) combine to show the statement at the 
start of the problem.
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\item{} {\it Solution:}  The desired statement follows from parts (d) and (e) and Theorem 3.3.4 (equality of the two one-sided limits implies the two-sided limit exists and equals the common value).  
\bye
