\magnification=\magstep1
\centerline{Mathematics 242 -- Principles of Analysis}
\centerline{Selected Solutions -- Problem Set 3}
\centerline{October 20, 2000}
\bigskip
\noindent
Section 3.1
\bigskip
\noindent 
3.  We are given that $|f(x)| \le 7x^2$ for all $x$.  This is equivalent
to the inequalities 
$$-7x^2 \le f(x) \le 7x^2\leqno(1)$$
for all $x$.  
\bigskip
\item{a.} Since $\lim_{x\to 0} -7x^2 = 0 = \lim_{x\to 0} 7x^2$,
the Limit Squeeze Theorem shows that the limit of $f(x)$ is
$\lim_{x\to 0} f(x) = 0$.
\item{b.} Here we have $\lim_{x\to 0} -7x^2 = -28$ but $\lim_{x\to 0} 7x^2 = 28$.
The Squeeze Theorem does not apply, and we cannot say more than the following:
{\it If the limit exists, then by Theorem 3.2.4, it is between -28 and +28}.
However, there is no guarantee the limit exists at all in this case.
$f(x)$ could have a jump discontinuity (or worse) at $x = 2$.
\bigskip
\noindent
9.  We are given that $\lim_{x\to c} f(x)$ exists.  Hence for all 
$\epsilon > 0$, there exist $\delta > 0$ such that 
$|f(x) - L| < \epsilon$ for all $x$ with $0 < |x - c| < \delta$.  
With $\epsilon = 1$, for instance, we get a corresponding $\delta$
such that $|f(x) - L| < 1$ for all $x$ in the deleted neighborhood 
$0 < |x - c| < \delta$ of $c$.  Here's a clever way to get an upper bound
on $|f(x)|$ using the triangle inequality (idea taken from Jackie 
Pournaras's solution):  For $x$ in this deleted neighborhood of $c$:
$$|f(x)| = |f(x) - L + L| \le |f(x) - L| + |L| < 1 + |L|.$$
Hence $M = 1 + |L|$ is one upper bound.
\bigskip
\noindent
10.  We are given that $\lim_{x\to c} f(x) = L$
\item{a.}  For all $\epsilon > 0$, there exist $\delta > 0$ such 
that $|f(x) - L| < \epsilon$ for all $x$ with $0 < |x - c| < \delta$.
But then for $x$ in the same deleted neighborhood of $c$, by a general
property of absolute values (Exercise 4, Section 1.3):
$$||f(x)| - |L|| \le |f(x) - L| < \epsilon.$$
Hence  $\lim_{x\to c} |f(x)| = |L|$.
\item{b.}  Factoring the difference of squares, we have 
$|f^2(x) - L^2| = |f(x) + L|\cdot |f(x) - L|$.
By problem 9, we know that there is some $\delta_1$ and $M > 0$ such
that $|f(x)| \le M$ for all $x$ with $0 < |x-c| < \delta_1$.  Hence, by 
the triangle inequality,
$|f(x) + L| \le |f(x)| + |L| \le M + |L|$ for all $x$ in the deleted
neighborhood $0 < |x-c| < \delta_1$.  Then since $\lim_{x\to c} f(x) = L$,
for any $\epsilon > 0$, there exists a $\delta_2$ such 
that $|f(x) - L| < {\epsilon\over M + |L|}$ for all $x$ with
$0 < |x - c| < \delta_2$.  But then with $\delta = \min(\delta_1,\delta_2)$,
we get for all $x$ with $0 < |x - c| < \delta$:
$$\eqalign{
|f^2(x) - L^2| &= |f(x) + L|\cdot |f(x) - L|\cr
               &\le (M + |L|)\cdot {\epsilon\over M + |L|}\cr
               &=\epsilon\cr}$$
Hence  $\lim_{x\to c} f^2(x) = L^2.$          
\bigskip
\noindent
Section 3.2
\noindent
7.  Given that $f(x) \ge 0$ on a deleted neighborhood
of $c$, and $\lim_{x\to c} f(x) = L$.
\item{a)}  Suppose $L = 0$.  Then for all $\epsilon > 0$, there
exists a $\delta > 0$ such that $|f(x)| < \epsilon^2$ for all
$x$ with $0 < |x-c| < \delta$.  But then, for $x$ in this
deleted neighborhood of $c$, 
$$|\sqrt{f(x)}| < \sqrt{\epsilon^2} = \epsilon,$$
which shows that $\lim_{x\to c} \sqrt{f(x)} = 0 = \sqrt{0}$.
\item{b)}  Now suppose that $L > 0$.  We have
$$|\sqrt{f(x)} - \sqrt{L}| = |\sqrt{f(x)} - \sqrt{L}| 
\cdot{|\sqrt{f(x)} + \sqrt{L}|\over |\sqrt{f(x)} + \sqrt{L}|} =
{|f(x) - L|\over |\sqrt{f(x)} + \sqrt{L}|} \le {|f(x) - L|\over \sqrt{L}}$$
where the last inequality follows since $\sqrt{f(x)} \ge 0$ for all $x$
in the deleted neighborhood of $c$. Then since $\lim_{x\to c} f(x) = L$,
for any $\epsilon > 0$, there exists a $\delta$ such 
that $|f(x) - L| < \epsilon\sqrt{L}$ for all $x$ with
$0 < |x - c| < \delta$.   For $x$ in this deleted neighborhood,
$$|\sqrt{f(x)} - \sqrt{L}|\le {|f(x) - L|\over \sqrt{L}} < 
{\epsilon \sqrt{L}\over \sqrt{L}} = \epsilon$$
Therefore, $\lim_{x\to c} \sqrt{f(x)} = \sqrt{L}$.
\bigskip
\noindent
8.  The idea here is to use the limit theorems from this section,
{\it not} the $\epsilon$ - $\delta$ definition of limits.
Given $\lim_{x\to c} f(x) = L$
\item{a.}  To show that $\lim_{x\to c} |f(x)| = |L|$, consider
the cases $L > 0, L = 0, L < 0$.  First, if $L > 0$, then
Theorem 3.2.2 says that there is a deleted neighborhood of $c$ where
$f(x) > 0$.  We only need to consider this neighborhood to compute
$\lim_{x\to c} |f(x)| = |L|$.  But on this neighborhood, 
$|f(x)| = f(x)$ and $|L| = L$, so there is really nothing to prove.
Next, if $L < 0$, then by the same theorem, there is a deleted 
neighborhood of $c$ where $f(x) < 0$, and we only need
consider $x$ in that neighborhood to compute the limit.  We have
by the limit product rule (Theorem 3.2.3) with one function the 
constant $-1$:
$$\eqalign{
\lim_{x\to c} |f(x)| &= \lim_{x\to c} -f(x)\cr
&=(\lim_{x\to c} -1)(\lim_{x\to c} f(x))\cr
&= (-1)(L)\cr
&= |L|\cr }$$
Finally, if $L = 0$, then $|f(x)|$ must also be going to 0 as
$x \to c$ (why?)
\item{b.}  Use the limit product rule (Theorem 3.2.3 b) with $g(x) = f(x)$:
$$\lim_{x\to c} f(x)^2 = (\lim_{x\to c} f(x))(\lim_{x\to c} f(x)) = L\cdot L = L^2$$
\bigskip
\noindent
Section 3.3
\bigskip
\noindent
5 c)  We must show that $\lim_{x \to 4^{-}} {2x + 3\over x - 4} = -\infty$.
By the definition, this means that we must show that for all $M > 0$
there exists a $\delta > 0$ such that
$${2x + 3\over x - 4} < -M\leqno(2)$$
for all $x$ in some half-neighborhood $4 - \delta < x < 4$.
Solving (2) for $x$, if $x < 4$, so $x - 4 < 0$ we get
$$\eqalign{
{2x + 3\over x - 4} < -M &\Leftrightarrow 2x + 3 > -(x - 4)M\cr
&\Leftrightarrow 2x + 3 > -Mx + 4M\cr
&\Leftrightarrow  (M + 2)x > 4M - 3\cr
&\Leftrightarrow x > {4M - 3\over M + 2}\cr
&\Leftrightarrow x > {4M + 8 - 11\over M+2}\cr
&\Leftrightarrow x > 4 - {11\over M + 2}\cr}$$
Hence if we take $\delta = {11\over M + 2} > 0$, then
$4 - \delta < x < 4$ implies ${2x + 3\over x - 4} < -M$, which is 
what we had to show.
\bye
        

 