\magnification=\magstep1
\centerline{Mathematics 242 -- Principles of Analysis}
\centerline{Selected Solutions -- Problem Set 6}
\centerline{November 8, 2004}
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\noindent
20.4 c)  We want to show that $\lim_{x\to 2} x^3 = 8$.  Note that 
$|x^3  - 8| = |x - 2|\cdot |x^2 + 2x + 4|$.  If $|x - 2| < 1$, for
instance, then $1 < x < 3$, so $7 < x^2 + 2x + 4 < 19$.  So we
are ready for the proof:

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\noindent
Let $\varepsilon > 0$, and let $\delta = \min(1, \varepsilon/19)$.  Then
for all $x$ with $0 < |x - 2| < \delta$, we have
$$| x^3 - 8 | = |x - 2|\cdot |x^2 + 2x + 4| < \varepsilon/19 \cdot 19 = \varepsilon.$$
Hence $\lim_{x \to 2} x^3 = 8$.

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\noindent
20.6 b) Given: $f(x) \ge 0$ for all $x \in D$ and $\lim_{x \to c} f(x) = L$.
Note first that since $\sqrt{f(x)} \ge 0$, 
$$\left|{1\over \sqrt{f(x)} + \sqrt{L}}\right| < {1\over \sqrt{L}},$$
for all $x\in D$.  Now let $\varepsilon > 0$
be arbitrary.  Since $\lim_{x \to c} f(x) = L$, there exists a $\delta > 0$
such that $|f(x) - L| < \varepsilon\cdot \sqrt{L}$ for all $x \in D$
with $|x - c| < \delta$.  But then if
$0 < |x - c| < \delta$, we have by the difference of squares factorization:
$$|\sqrt{f(x)} - \sqrt{L}| = |f(x) - L|\cdot \left|{1\over \sqrt{f(x)} + \sqrt{L}}\right|
< \varepsilon\cdot \sqrt{L} \cdot {1\over \sqrt{L}} = \varepsilon.$$
This shows $\lim_{x \to c} \sqrt{f(x)} = \sqrt{L}$.
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\noindent
20.7 b)  The function $f(x) = \sin(1/x)$ has no limit as $x \to 0^+$ because there exist
sequences $s_n$ converging to zero for which the sequences of values $f(x)$ 
have different limits.  For instance, with $s_n = {1\over n\pi}$ we have
$f(s_n)= \sin(n\pi) = 0$ all $n$.  Hence $\lim_{n\to\infty} f(s_n) = 0$.
But for $t_n = {2\over (4n+1)\pi}$, we have $\sin(t_n) = \sin(2n\pi + \pi/2) = 1$
for all $n$.  Hence $\lim_{n\to\infty}f(t_n) = 1$.  By Theorem 20.8, $\lim_{x\to 0^+}f(x)$
does not exist.

\noindent
c) $\lim_{x \to 0^+} x \sin(1/x) = 0$.  Proof: Let $\varepsilon > 0$, and take
$\delta = \varepsilon$.  Then for all $x > 0$ with $0 < |x - 0| < \delta = \varepsilon$,
$$|x\sin(1/x)| = |x| |\sin(1/x)| \le |x|\cdot 1 < |x| < \varepsilon.$$
(We are using the fact that $|\sin(u)| \le 1$ for all $u$ here.)  

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\noindent
20.8 a)  We want to show that if $\lim_{x\to c}f(x) = L$, then the limit
$L$ is unique, first using the definition.  Suppose $\lim_{x\to c}f(x) = L$
and $\lim_{x\to c}f(x) = M$ are both true.  Then for all $\varepsilon > 0$,
there exist $\delta_1 > 0$ such that $|f(x) - L| < \varepsilon/2$ for 
all $x\in D$ with $0 < |x - c| < \delta_1$ and similarly there exists
$\delta_2$ such that $|f(x) - M| < \varepsilon/2$ for all $x\in D$
with $0 < |x - c| < \delta_2$.  Then for $\delta=\min(\delta_1,\delta_2)$
and all $x \in D$ with $0 < |x - c| < \delta$  
$$|L - M| = |L - f(x) + f(x) - M| \le |f(x) - L| + |f(x) - M| < \
\varepsilon/2 + \varepsilon/2 = \varepsilon$$
Since $0 \le |L - M| < \varepsilon$ for all positive $\varepsilon$, we
must have $|L - M| = 0$, so $L = M$.

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\noindent
b) Now we prove the same result a different way, using the ``sequential
criterion'' for limits (Theorem 20.8), and the uniqueness for
limits of sequences that we proved earlier.  We assume that $\lim_{x\to c} f(x) = L$
and $ \lim_{x\to c} f(x) = L$ as in part a.  Then by Theorem 20.8, for
any sequence $s_n$ in $D$ with $s_n \ne c$ for all $n$
and $\lim_{n\to\infty} s_n = c$, we
have that $f(s_n)$ is convergent to the limit of $f(x)$.  Thus we have
have $\lim_{n\to\infty}f(s_n) = L$ and $\lim_{n\to\infty}f(s_n) = M$.
But Theorem 16.14, proved before, says:  The limit of any convergent
sequence is unique.  So $L = M$.
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\noindent
20.11.  The idea for this one is like the proof in 8 b) above.  We assume
that $f(x) \le g(x) \le h(x)$ for all $x \in D$, and that
$\lim_{x\to c} f(x) = L = \lim_{x\to c}h(x)$.  Then Theorem 20.8
says:  If $s_n$ is any sequence in $D$ converging to $c$ (with $s_n \ne c$ for all $n$), we have
$\lim_{n\to\infty}f(s_n) = \lim_{n\to\infty} h(s_n) = L$.  But now
we use the result from Theorem 17.4 (proved previously):
If $t_n \le u_n$ for all $n$, and both
sequences are convergent, then $\lim_{n\to\infty}t_n \le \lim_{n\to\infty} u_n$.
As in Exercise 17.20, since $f(s_n) \le g(s_n) \le h(s_n)$ for all $n$,
we have that $g(s_n)$ must be convergent to $L$ as well.  (Proof:
Since $\lim_{n\to\infty} f(s_n) = \lim_{n\to\infty} h(s_n)$, for all $\varepsilon > 0$,
there exists $N$ such that $|f(s_n) - L| < \varepsilon$ and 
$|h(s_n) - L| < \varepsilon$ for all $n \ge N$.  This says
$L - \varepsilon < f(s_n) \le g(s_n) \le h(s_n) < L + \varepsilon$
for all $n \ge N$. So $|g(s_n) - L| < \varepsilon$ also.
Since this is true for 
all sequences in $D$ converging to $c$, $\lim_{x\to c}g(x) = L$ by
Theorem 20.8.
\bye
