Exercise 1.5: Solution

Suppose $\delta >0$. If $\vert x-a\vert<\epsilon /2$ and $\vert y-b\vert<\epsilon /2$, then
$\vert(x+y)-(a+b)\vert<\epsilon$.

If $x$ is close enough to $a$ and $y$ is close enough to $b$ then $x+y$ is TOLERABLY close to $a+b$.

Proof

$\begin{array}{rcccl} -\frac{\epsilon }{2} & < & x-a & < \frac{\epsilon }{2} \\ -\frac{\epsilon }{2} & < & y-b & < \frac{\epsilon }{2} \end{array}$

this gives, by addition:

$\begin{array}{rcccl} -\epsilon & < & (x+y)-(a+b) & < & \epsilon \\ & & \vert(x+y)-(a+b)\vert & < & \epsilon . \end{array}$

Exercise 1.8: Solution

If $\vert x\vert<\epsilon /3$ and $\vert y\vert<\epsilon /3$ and $\vert z\vert<\epsilon /3$, then $\vert x+y+z\vert<\epsilon$.

If $x$, $y$, and $z$ are VERY VERY small then $x+y+z$ is TOLERABLY small.

Proof

$$\begin{eqnarray*} \vert x+y+z\vert & \leq & \vert x+y\vert+\vert z\vert \\ &&\\... ...rac{\epsilon }{3}+\frac{\epsilon }{3}\\ &&\\ & = & \epsilon . \end{eqnarray*}$$

The distance from $x+y+z$ to 0 is $\leq$ the distance from $x+y$ to 0 plus the distance from $z$ to 0, and this is $\leq$ the distance from $x$ to 0 plus the distance from $y$ to 0 plus the distance from $z$ to 0, which are all $<$ one third of the TOLERANCE.

Exercise 1.9: Solution

If $\vert x\vert<\epsilon /2$ and $\vert y\vert<\epsilon /2$, then $\vert x-y\vert<\epsilon$.

If $x$ and $y$ are VERY small then $x-y=x+(-y)$ is TOLERABLY small.

Proof

$$\begin{eqnarray*} \vert x-y\vert & = & \vert x+(-y)\vert \\ & \leq & \vert x\v... ...< & \frac{\epsilon }{2}+\frac{\epsilon }{2}\\ & = & \epsilon . \end{eqnarray*}$$

The distance from $x-y$ to 0 is $\leq$ the distance from $x$ to 0 plus the distance from $-y$ to 0. This is $=$ to the distance from $x$ to 0 plus the distance from $y$ to 0, which are both $<$ one half of the TOLERANCE.

Exercise 1.10: Solution

If $\vert x-y\vert<\epsilon /2$ and $\vert y-z\vert<\epsilon /2$
then $\vert x-z\vert<\epsilon$.

If $x$ is VERY close to $y$ and $y$ is VERY close to $z$ then $x$ is TOLERABLY close to $z$.

Proof

$$\begin{eqnarray*} \vert x-z\vert & \leq & \vert x-y\vert+\vert y-z\vert \\ &&\\... ...< & \frac{\epsilon }{2}+\frac{\epsilon }{2}\\ & = & \epsilon . \end{eqnarray*}$$

The distance from $x$ to $z$ is $\leq$ the distance from $x$ to $y$ plus the distance from $y$ to $z$.
Both of these are $<$ one half the TOLERANCE.

Exercise 1.11a: Solution

If $\vert x-y\vert<\epsilon$ then
$\vert\vert x\vert-\vert y\vert\vert<\epsilon .$

If $x$ is TOLERABLY close to $y$ then $\vert x\vert$ is TOLERABLY close to $\vert y\vert$, that is, the distance from $x$ to 0 is TOLERABLY close to the distance from $y$ to 0.

Proof

Following the hint, it is sufficient to show that $\vert\vert a\vert-\vert b\vert\vert \leq \vert a-b\vert$. We have:

$$\begin{eqnarray*} \vert a\vert & = & \vert(a-b)+b\vert \\ & \leq & \vert a-b\vert+\vert b\vert, \end{eqnarray*}$$

so,

$$\begin{eqnarray*} \vert a\vert-\vert b\vert & \leq & \vert a-b\vert. \end{eqnarray*}$$

Also,

$$\begin{eqnarray*} \vert b\vert & = & \vert(b-a)+a\vert \\ & \leq & \vert b-a\vert+\vert a\vert, \end{eqnarray*}$$

so,

$$\begin{eqnarray*} -\vert a-b\vert & \leq & \vert a\vert-\vert b\vert. \end{eqnarray*}$$

This gives:

$$-\vert a-b\vert \leq \vert a\vert-\vert b\vert \leq \vert a-b\vert$$

$$\vert\vert a\vert-\vert b\vert\vert \leq \vert a-b\vert$$

Exercise 1.11b: Solution

If $\vert x\vert<3$ then
$\vert x^2+5x+4\vert<28.$

If the distance from $x$ to 0 is $< 3$ then the distance from $x^2+5x+4$ to 0 is $< 28.$

Proof

$$\begin{eqnarray*} \vert x^2+5x+4\vert & \leq & \vert x^2\vert+\vert 5x\vert+\ver... ...rt^2+5\vert x\vert+4\\ &&\\ & < & 3^2+5\cdot3+4\\ & = & 28. \end{eqnarray*}$$

The distance from $x^2+5x+4$ to 0 is $\leq$ the distance from $x^2$ to 0 plus the distance from $5x$ to 0 plus the distance from 4 to 0.
Since $x$ is $< 3$ this sum is $< 28.$

Exercise 1.12a: Solution

For some $\delta >0$ if $\vert x\vert<\delta$, then $\vert 2+x\vert<3.$

If $x$ is small enough then the distance from $x$ to 2 is $< 3$.

Proof

We may let $\delta=1.$
Indeed, if $\vert x\vert<1$, then

$$\begin{eqnarray*} \vert x+2\vert & = & \vert x-(-2)\vert\\ & \leq & \vert x-0\... ...t\\ & = & \vert x\vert+\vert 2\vert\\ & < & 1+2\\ & = & 3. \end{eqnarray*}$$

If $x$ is small enough the distance from $x$ to 0 is $<1$.


So the distance from $x$ to $-2$ is $\leq$ the distance from $x$ to 0 plus the distance from 0 to $-2$ which is $<1+2=3.$

Exercise 1.12b: Solution

For some $\delta >0$ if $\vert x-2\vert<\delta$, then $\vert x-3\vert<5/4.$

If $x$ is close enough to 2, then the distance from $x$ to 3 is $<5/4$.

Proof

We may let $\delta=1/4.$
Indeed, say $\vert x-2\vert<1/4$, then

$$\begin{eqnarray*} \vert x-3\vert & = & \vert x-2+2-3\vert\\ & \leq & \vert x-2\vert+\vert 2-3\vert\\ & < & 1/4+1\\ & = & 5/4. \end{eqnarray*}$$

If $x$ is close enough to 2 the distance from $x$ to 2 is $<1/4$.


So the distance from $x$ to $3$ is $\leq$ the distance from $x$ to 2 plus the distance from 2 to $3$ which is $<1/4+1=5/4.$

Exercise 1.13: Solution

For some $\delta >0$ if $\vert x\vert<\delta$, then $a-x>0.$

If $x$ is small enough then the displacement from $x$ to $a$ is positive.

Proof

We may let $\delta=a/2.$
Indeed, say $\vert x\vert<a/2$, then

$$-\frac{a}{2} < x < \frac{a}{2}$$

so$x<a$ and$a-x>0$.

If $x$ is small enough $x$ lies between $-a/2$ and $a/2$.




So $x$ is to the left of $a$ and the displacement from $x$ to $a$ is positive.

Exercise 1.14: Solution

For some $\delta >0$ if $\vert x-a\vert<\delta$, then $a-1<x<a-1.$

If $x$ is close enough to $a$, then $x$ lies between $a-1$ and $a+1$.

Proof

We may let $\delta=1.$
Indeed, if $\vert x-a\vert<1$, then


$-1 < x-a < 1$ so $a-1 < x < a+1.$

Precision=1.

Exercise 1.15: Solution

For some $\delta >0$ if $\vert x-a\vert<\delta$, then $a/2<x<3a/2.$

If $x$ is close enough to $a$, then $x$ lies between $a/2$ and $3a/2$.

Proof

We may let $\delta=a/2.$
Indeed, if $\vert x-a\vert<a/2$, then


$-a/2 < x-a < a/2$

so $a/2 < x < 3a/2$

Precision=$a/2$.

Exercise 1.16: Solution

For some $\delta >0$ if $\vert x-a\vert<\delta$, then $x>0$.

If $x$ is close enough to $a$, then $x$ is positive.

Proof

This follows from Exercise 1.15, since $x>a/2 \Rightarrow x>0$. Exercise 1.17: Solution

Say $\epsilon >0$. For some $\delta >0$, if $\vert x\vert<\delta$ then $x^2<\epsilon$.

If $x$ is small enough, $x^2$ is TOLERABLY small.

Proof

We may let $\delta=\sqrt\epsilon$.
Indeed, if $\vert x\vert<\sqrt\epsilon$ then

$$\begin{eqnarray*} x^2 & = & \vert x\vert^2 \\ & < & (\sqrt\epsilon )^2 \\ & = & \epsilon . \end{eqnarray*}$$

Precision= $\sqrt\epsilon .$

Exercise 1.18: Solution

Note that $\vert x+2\vert=\vert x-(-2)\vert$.
For some $\delta >0$, if $\vert x-2\vert<\delta$ then $\vert x+2\vert<5$.

If $x$ is close enough to 2, then the distance from $x$ to -2 is $<5$.

Proof

We may let $\delta=1$.
Indeed, say $\vert x-2\vert<1$. Then

$$\begin{eqnarray*} \vert x+2\vert & = & \vert x-(-2)\vert \\ & \leq & \vert x-2\vert+\vert 2-(-2)\vert \\ & < & 1+4 \\ & = & 5. \end{eqnarray*}$$

If $x$ is close enough to 2, then the distance from $x$ to 2 is $<1$. So, the distance from $x$ to -2 is $\leq$ the distance from $x$ to 2 plus the distance from 2 to -2.
This is $<1+4=5$.

Exercise 1.19: Solution

TOLERANCE=1
2 wires 5 cm., precision=1/24.
2 wires 8 cm., precision=1/15.

$\delta_1$=min( $\epsilon /24,1$)
$\delta_2$=$\epsilon /15$
TOLERANCE=25
2 wires 5 cm., precision=1.
2 wires 8 cm., precision=1/15.


Exercise 1.20: Solution

>From the picture on page 27, we can see that for the demand for A to be VERY VERY small requires $\vert x-5\vert<\epsilon /24$; the demand for B to be VERY VERY small requires $\vert x-5\vert<1$ and $\vert y-8\vert<\epsilon /3$; and the demand for C to be VERY VERY small requires $\vert y-8\vert<\epsilon /15.$ Asking $\vert x-5\vert<1$ is probably much less demanding than asking $\vert x-5\vert<\epsilon /24$; asking $\vert y-8\vert<\epsilon /3$ is much less demanding than asking $\vert y-8\vert<\epsilon /15$. Therefore, making B VERY VERY small is easier than making either A or C VERY VERY small.

Exercise 1.21: Solution

$$\left\{ \begin{minipage}[c]{2 cm} The area of the ideal recta... ...egin{minipage}[c]{1 cm} The area of B. \end{minipage}\right\}$$

$$% latex2html id marker 1230 \makebox[2.8 cm][r] = \left\{ ... ...egin{minipage}[c]{1 cm} $y-8$\ by $5-x$\end{minipage}\right\}$$

The difference between the areas of the ideal and practical rectangles is equal to {Area of A} minus {Area of B} plus {Area of C}. Each of these are VERY VERY small. By the triangle inequality:

$$\begin{eqnarray*} \vert x-y+z\vert & = & \vert x+(-y)+z\vert\\ & \leq & \vert ... ...t+\vert z\vert\\ & = & \vert x\vert+\vert y\vert+\vert z\vert. \end{eqnarray*}$$

So the difference between the areas is TOLERABLY small.
Exercise 1.22: Solution

	Area of A	$=$	$x-2$ by 2.
	Area of B	$=$	$x-2$ by $x-2$.
	Area of C	$=$	$x-2$ by 2.

$$x^2-4=2(x-2)+(x-2)^2+2(x-2)$$

If $\vert x-2\vert<\epsilon /6$ then $2\vert x-2\vert<\epsilon /3$.
If $\vert x-2\vert<\sqrt{\epsilon /3}$ then $(x-2)^2<\epsilon /3$.
If $\vert x-2\vert<\epsilon /6$ then $2\vert x-2\vert<\epsilon /3$.

If $x$ is close enough to 2 then the areas of rectangles A, B and C are VERY VERY small.

Say $x-2<\min(\epsilon /6,\sqrt{\epsilon /3})=\delta$, then $\vert 2(x-2)\vert<\epsilon /3$ and $(x-2)^2<\epsilon /3$, so:

$$\begin{eqnarray*} \vert x^2-4\vert & = & \vert 2(x-2)+2(x-2)+(x-2)^2\vert \\ &... ...{3}+\frac{\epsilon }{3}+\frac{\epsilon }{3}\\ & = & \epsilon . \end{eqnarray*}$$

Note that we used:

$$\begin{eqnarray*} x^2-4 & = & 2(x-2)+2(x-2)+(x-2)^2 \\ & = & (x-2)(2+2+x-2) \\ & = & (x-2)(x+2) \end{eqnarray*}$$

as explained in greater detail in the next section.
Exercise 1.23: Solution

$$\begin{eqnarray*} xyz & = & [(x-4)+4][(y-5)+5][(z-6)+6] \\ & = & 120+30(x-4)+2... ... & (x-4)(y-5)6+(x-4)5(z-6)+4(y-5)(z-6)+\\ & & (x-4)(y-5)(z-6). \end{eqnarray*}$$

There are seven things which each need to be made VERY VERY VERY VERY VERY VERY VERY small. We need:

$\begin{array}{ccccll} \vert x-4\vert & < & \frac{\e... ...ert y-5\vert<1$} \\ &&&&& \mbox{and $\vert z-6\vert<\epsilon /7$.} \end{array}$

Manufacturing should be instructed to make the following:

	4 wires @ 4 cm	precision= $\min(\epsilon /210,1)$
	4 wires @ 5 cm	precision= $\min(\epsilon /168,1)$
	4 wires @ 6 cm	precision=$\epsilon /140$.

Exercise 1.24: Solution

$$\begin{eqnarray*} \vert x^2-4\vert & = & \vert x-2\vert\vert x+2\vert\\ \delta & = & \min(1,\epsilon /5) \end{eqnarray*}$$

a) $\:\epsilon =.1$

$$\begin{eqnarray*} \delta_1 & = & \min(1,.1/5)\\ & = & \frac{1}{50}. \end{eqnarray*}$$

b)$\:\epsilon =6$

$$\begin{eqnarray*} \delta_2 & = & \min(1,6/5)\\ & = & 1. \end{eqnarray*}$$

Exercise 1.25: Solution

Let $a$ be a number. Say $\epsilon >0$. There exists $\delta >0$ such that if $\vert x-a\vert<\delta$ then $\vert x^2-a^2\vert<\epsilon .$

If $x$ is close enough to $a$, then $x^2$ is TOLERABLY close to $a^2$.

For the proof use:

$$\begin{eqnarray*} \vert x^2-a^2\vert & = & \vert(x-a)(x+a)\vert \\ & = & \vert x-a\vert\vert x+a\vert. \end{eqnarray*}$$

Concept:

$\begin{array}{ccc} x \mbox{ close enough to } a & \Rightarrow & \vert x+a\vert... ...\\ \vert x-a\vert<1 & \Rightarrow & \vert x+a\vert<2\vert a\vert+1 \end{array}$

We also want:

$\vert x-a\vert<\frac{\epsilon }{2\vert a\vert+1}$.

Proof

Say $\delta=\min(1,\epsilon /(2\vert a\vert+1))$. Suppose $\vert x-a\vert<\delta$.
Then $\vert x-a\vert<\epsilon /(2\vert a\vert+1)$.
Also:

$$\begin{eqnarray*} \vert x+a\vert & = & \vert x-(-a)\vert\\ & \leq & \vert x-a\... ...a-(-a)\vert\\ & < & 1+\vert 2a\vert\\ & = & 2\vert a\vert+1. \end{eqnarray*}$$

Thus:

If $x$ is close enough$^1$ to $a$ then
$\vert x-a\vert<\epsilon /(2\vert a\vert+1)$.

If $x$ is close enough$^2$ to $a$ then
$\vert x+a\vert<2\vert a\vert+1$.

So if $x$ is close enough to $a$, both happen.
Thus:

$$\begin{eqnarray*} \vert x^2-a^2\vert & = & \vert x-a\vert\vert x+a\vert\\ & < ... ...{2\vert a\vert+1} \right) (2\vert a\vert+1)\\ & = & \epsilon . \end{eqnarray*}$$

Exercise 1.26: Solution

Say $a\neq 0$. Say $\epsilon >0$. There exists $\delta >0$ such that if $\vert x-a\vert<\delta$ then $\vert 1/x-1/a\vert<\epsilon .$

If $x$ is close enough to $a$, then $1/x$ is TOLERABLY close to $1/a$.

Note:

$\begin{array}{ccc} \vert x-a\vert<\frac{\vert a\vert}{2} & \Rightarrow & \vert x\vert>\frac{\vert a\vert}{2}. \end{array}$

Proof

$$\begin{eqnarray*} \vert a\vert & = & \vert a-0\vert\\ & \leq & \vert x-0\vert+... ...rt+\vert a-x\vert\\ & < & \vert x\vert+\frac{\vert a\vert}{2}. \end{eqnarray*}$$

Thus:

The distance from 0 to $a$ is $\leq$ the distance from 0 to $x$ plus the distance from $x$ to $a$ which is $<$ $\vert x\vert+\vert a\vert/2$.
Thus:

$$\begin{eqnarray*} \vert a\vert & < & \vert x\vert+\frac{\vert a\vert}{2}\\ \vert x\vert & > & \frac{\vert a\vert}{2}. \end{eqnarray*}$$

Proof

Say $\delta=\min(\epsilon a^2/2,\vert a\vert/2)$.
Suppose $\vert x-a\vert<\delta$. Then
$\vert x-a\vert<\vert a\vert/2 \Rightarrow \vert x\vert>\vert a\vert/2$, so

$$\begin{eqnarray*} \frac{1}{\vert a\vert\vert x\vert} & < & \frac{2}{a^2}. \end{eqnarray*}$$

Also:

$$\begin{eqnarray*} \vert x-a\vert & < & \frac{\epsilon a^2}{2}, \end{eqnarray*}$$

Thus:

If $x$ is close enough to $a$ then
$\vert x\vert>\vert a\vert/2$, so

$$\begin{eqnarray*} \frac{1}{\vert a\vert\vert x\vert} & < & \frac{2}{a^2}. \end{eqnarray*}$$

Thus:

[ Exercise 1.26: Solution continued ]

$$\begin{eqnarray*} \left\vert\frac{1}{x}-\frac{1}{a}\right\vert & = & \left\vert\... ...psilon \cdot\frac{a^2}{2}\cdot\frac{2}{a^2}\\ & = & \epsilon . \end{eqnarray*}$$

Exercise 1.27: Solution

Say $\epsilon >0$. There exists $\delta >0$ such that if $\vert x-2\vert<\delta$ then
$\vert x^3-8\vert<\epsilon .$

If $x$ is close enough to 2, then $x^3$ is TOLERABLY close to 8.

Note:

$$\begin{eqnarray*} \vert x^3-8\vert & = & \vert(x-2)(x^2+2x+4)\vert\\ & = & \vert x-2\vert\vert x^2+2x+4\vert \hspace{1 cm} (*). \end{eqnarray*}$$

If $x$ is close enough to 2 then $\vert x\vert<3$.
$\Downarrow$
$\vert x^2+2x+4\vert<9+6+4=19.$
$\Downarrow$
We need $\vert x-2\vert<\frac{\epsilon }{19}.$
$\Downarrow$
Let $\delta=\min(\epsilon /19,1)$

[ continued next page ]
[ Exercise 1.27: Solution continued ]

Proof

Say $\delta=\min(\epsilon /19,1)$.
Suppose $\vert x-2\vert<\delta$. Then
$\vert x-2\vert<1$, so

$$\begin{eqnarray*} \vert x-0\vert & \leq & \vert x-2\vert+\vert 2-0\vert\\ \vert x\vert & < & 1+2=3. \end{eqnarray*}$$

So,

$$\begin{eqnarray*} \vert x^2+2x+4\vert & \leq & \vert x^2\vert+\vert 2x\vert+\vert 4\vert\\ & < & 9+6+4=19. \end{eqnarray*}$$

And,

$$\begin{eqnarray*} \vert x-2\vert & < & \frac{\epsilon }{19}. \end{eqnarray*}$$

Therefore:

If $x$ is close enough$^1$ to 2


then $\vert x\vert<3$,




so $\vert x^2+2x+4\vert<19$.



If $x$ is close enough$^2$ to 2 then
$\vert x-2\vert<\epsilon /19.$

If $x$ is close enough$^3$ to 2 then
BOTH HAPPEN.
Therefore:

$$\begin{eqnarray*} \vert x^3-8\vert & < & \frac{\epsilon }{19}\cdot 19 = \epsilon . \end{eqnarray*}$$

Using (*) above. Exercise 1.28: Solution

Say $\epsilon >0$. There exists $\delta >0$ such that if $\vert x-3\vert<\delta$ then
$\vert(x^2-2x+3)-6\vert<\epsilon .$

If $x$ is close enough to 3, then
$x^2-2x+3$ is TOLERABLY close to 6.

Note:

$$\begin{eqnarray*} \vert x^2-2x+3-6\vert & = & \vert x^2-2x-3\vert\\ & = & \vert x-3\vert\vert x+1\vert. \end{eqnarray*}$$

$x$ is close enough to 3.
$\Downarrow$
$\vert x+1\vert<5$ AND $\vert x-3\vert<\epsilon /5$.
$\Downarrow$
Let $\delta=\min(\epsilon /5,1)$

[ continued next page ]
[ Exercise 1.28: Solution continued ]

Proof

Say $\delta=\min(\epsilon /5,1)$.
Suppose $\vert x-3\vert<\delta$. Then
$\vert x-3\vert<1$, so

$$\begin{eqnarray*} \vert x+1\vert & = & \vert x-(-1)\vert\\ & \leq & \vert x-3\vert+\vert 3-(-1)\vert\\ & < & 1+4=5 \end{eqnarray*}$$

Also,

$$\begin{eqnarray*} \vert x-3\vert & < & \frac{\epsilon }{5}. \end{eqnarray*}$$

Therefore:

If $x$ is close enough$^1$ to 3



then $\vert x+1\vert<5$,




If $x$ is close enough$^2$ to 3 then
$\vert x-3\vert<\epsilon /5.$

If $x$ is close enough$^3$ to 3 then
BOTH HAPPEN.
Therefore:

$$\begin{eqnarray*} \vert x^2-2x+3-6\vert & = & \vert x-3\vert\vert x+1\vert\\ & < & \frac{\epsilon }{5})\cdot 5 = \epsilon . \end{eqnarray*}$$

Exercise 1.29: Solution

Say $\epsilon >0$. There exists $\delta >0$ such that if $\vert x-4\vert<\delta$ then
$\vert\sqrt{x}-2\vert<\epsilon .$

If $x$ is close enough to 4, then
$\sqrt{x}$ is TOLERABLY close to 2.

Note:

$$\begin{eqnarray*} \vert\sqrt{x}-2\vert & = & \left\vert\frac{(\sqrt{x}-2)(\sqrt{... ...rt{x}+2}\right\vert\\ & = & \vert x-4\vert\frac{1}{\sqrt{x}+2} \end{eqnarray*}$$

$$\begin{eqnarray*} % latex2html id marker 1182\mbox{$x$\ is close enough to 4} ... ...silon \\ & \Rightarrow & \mbox{ Let }\delta=\min(3,3\epsilon ) \end{eqnarray*}$$

[ continued next page ]
[ Exercise 1.29: Solution continued ]

Proof

Say $\delta=\min(3,3\epsilon )$.
Suppose $\vert x-4\vert<\delta$. Then
$\vert x-4\vert<3$, or

$\begin{array}{ccccc} -3 & < & x-4 & < & 3\\ 1 & < & x & < & 7 \end{array}$

So,

$$\begin{eqnarray*} \vert x\vert & > & 1 \\ \sqrt{x} & > & 1\\ \sqrt{x}+2 & > & 3\\ \frac{1}{\sqrt{x}+2} & < & \frac{1}{3}. \end{eqnarray*}$$

Also $\vert x-4\vert<3\epsilon$.


Therefore:

If $x$ is close enough$^1$ to 4








then $\sqrt{x}>1$,
so $\sqrt{x}+2>3$


and $1/(\sqrt{x}+2)<1/3$.





If $x$ is close enough$^2$ to 4 then
$\vert x-4\vert<3\epsilon .$

Therefore:

$$\begin{eqnarray*} \vert\sqrt{x}-2\vert & = & \vert x-4\vert\frac{1}{\sqrt{x}+2}\\ & < & 3\epsilon \cdot \frac{1}{3} = \epsilon . \end{eqnarray*}$$

Exercise 1.30: Solution

The customer wants the resistance to be about 6 ohms, so that the current will be about 20 amps. His TOLERANCE for the current is 1 amp.

If $x$ is close enough to 6 then $120/x$ is TOLERABLY close to 20, TOLERANCE=1.

We need:

$$\begin{eqnarray*} \left\vert\frac{120}{x}-20\right\vert<1 & \Leftrightarrow & \... ...ow & \left\vert\frac{1}{x}-\frac{1}{6}\right\vert<\frac{1}{120}. \end{eqnarray*}$$

This is the same as $1/x$ is TOLERABLY close to 1/6, TOLERANCE=1/120. >From Problem 26 with $a=6$ and $\epsilon =1/120$:

$$\begin{eqnarray*} \delta & = & \min\left(\frac{\epsilon a^2}{2},\frac{\vert a\vert}{2}\right)\\ & = & \frac{1}{120}\cdot 18=\frac{3}{20}. \end{eqnarray*}$$

`$\Box$

Exercise 1.31: Solution

Say $\epsilon >0.$ For some $\delta >0$, if $\vert x\vert<\delta$, then $\vert x\sin x\vert<\epsilon$.

If $x$ is small enough then $\vert x\sin x\vert$ is TOLERABLY small.

Plan:

$g(x)=\frac{\vert x\sin x\vert}{\vert x\vert}=\vert\sin x\vert \hspace{1 cm} x\neq0.$

We may as well let $g(x)=\vert\sin x\vert$.

If $\vert x\vert<1$ then $\vert\sin x\vert\leq 1$.

Put $\delta=\min(1,\epsilon )$.
Say $\vert x\vert<\delta$. Then
$\vert x\vert<1\Rightarrow \vert\sin x\vert\leq1$, and
$\vert x\vert<\epsilon$,

thus:

If $x$ is small enough$^1$ then $\vert\sin x\vert\leq 1$.

If $x$ is small enough$^2$,

then $\vert x\vert<\epsilon .$
So if $x$ is small enough$^3$ both happen, thus:

$\vert x \sin x\vert=\vert x\vert\vert\sin x\vert<\epsilon \cdot 1=\epsilon .$

`$\Box$

Exercise 1.32: Solution

Say $\epsilon >0.$ For some $\delta >0$, if $\vert x\vert<\delta$, then $\vert x\vert^{1/2}\vert\cos x\vert<\epsilon$.

If $x$ is small enough, $\vert x\vert^{1/2}\vert\cos x\vert$ is TOLERABLY small.

Plan:

$g(x)=\frac{\vert x\vert^{1/2}\vert\cos x\vert}{\vert x\vert}= \frac{\vert\cos x\vert}{\vert x\vert^{1/2}}\leq M \hspace{1 cm}\mbox{???}$

Change plans.
Concept:

$\begin{array}{cccl} \vert\cos x\vert & \leq & 1 & \hspace{1 cm} \mbox{Always}.\... ...& < & \epsilon & \hspace{1 cm} \mbox{If } \vert x\vert<\epsilon ^2. \end{array}$

Let $\delta=\epsilon ^2$.
Say $\vert x\vert<\delta$. Then $\vert x\vert^{1/2}<\epsilon$ so:

$\vert x\vert^{1/2}\vert\cos x\vert<\epsilon \cdot 1=\epsilon .$

`$\Box$

Exercise 1.33: Solution

$\begin{array}{lcccc} \mbox{We want: } & & \vert 10f(x)-.84\vert & < & .02\\ &... ...vert & < & .02\\ & \mbox{i.e.} & \vert f(x)-.084\vert & < & .002. \end{array}$

``Frank, with what precision $\delta$ must we manufacture these wires to be sure that $\vert f(x)-.084\vert<.002$ ?''

10 wires, each with 20% copper at a precision of $\delta$, will work.

According to Frank,
if $\vert x-20\vert<\delta$,
then $\vert f(x)-.084\vert<.002.$

$\Rightarrow \vert 10f(x)-.84\vert<.02.$

If $x$ is close enough to 20 then $f(x)$ is tolerably close to .084 (tolerance=.002), and $10f(x)$ is TOLERABLYclose to .84 (TOLERANCE=.02).

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Exercise 1.34: Solution

Concept:

$L=.084\hspace{.6 cm}M=.056\hspace{.6 cm}\epsilon =.00003$

$$\begin{eqnarray*} f(x)g(x)-LM & = & (f(x)-L)(g(x)-M)\\ & & +L(g(x)-M)\\ & & +M(f(x)-L). \end{eqnarray*}$$

Each term on the right must be VERY VERY small. That is, we must try to get: $\vspace{-.8 cm}$

$$\begin{eqnarray*} \vert g(x)-M\vert & < & \frac{\epsilon }{3L}\\ \vert f(x)-L\v... ...)-L\vert<1 & \mbox{AND} & \vert g(x)-M\vert<\frac{\epsilon }{3}. \end{eqnarray*}$$

``Frank, with what precision $\delta_1$ must we manufacture these wires so that $\vert f(x)-.084\vert<\min(.00001/.056,1)=.00001/.056?$''
``George, with what precision $\delta_2$ must we manufacture these wires so that $\vert g(x)-.056\vert<\min(.00001/.084,.00001)=.00001?$''
Message to manufacturing: 2 wires, each with 20% copper, precision= $\min(\delta_1,\delta_2)$.

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Exercise 1.35: Solution

Let $h(r)=1/r$. Then $1/f(x)=h(f(x)).$ >From Exercise 26 we get, for Harriet, $\eta=\min(\epsilon a^2/2,\vert a\vert/2).$ If $\vert r-a\vert<\eta$ then $\vert 1/r-1/a\vert<\epsilon$. That is, if $r$ is close enough to $a$ (precision $\eta$) then $1/r$ is TOLERABLY close to $1/a$ (TOLERANCE=$\epsilon$).

In our case, $a=.084$, and $\epsilon =3/84$, so

$$\begin{eqnarray*} \eta & = & \min\left(\frac{3}{84}\frac{(.084)^2}{2},\frac{.084}{2}\right)\\ & = & \frac{3}{84}\frac{(.084)^2}{2}=.000126. \end{eqnarray*}$$

>From Exercise 26 we know that if $\vert r-.084\vert<.000126$ then

$$\begin{eqnarray*} \left\vert\frac{1}{r}-\frac{1}{.084}\right\vert & < & \frac{3}{84}. \end{eqnarray*}$$

Frank can give us a precision $\delta$ such that

$$\begin{eqnarray*} \vert x-20\vert<\delta & \Rightarrow & \vert f(x)-.084\vert<.0... ...\left\vert\frac{1}{f(x)}-\frac{1}{.084}\right\vert<\frac{3}{84}. \end{eqnarray*}$$

``Frank, with what precision $\delta$ must we manufacture the wire so that $f(x)$, its resistance, is tolerably close to .084 (tolerance=.000126)?''
Message to manufacturing: 1 wire with 20% copper, precision=$\delta$.

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Exercise 1.36: Solution

We have that $1/r=1/f(x)+1/g(x)$. We want $r$ to be TOLERABLY close to 4704/140000 (TOLERANCE=10/140000).
Note that

$$\begin{eqnarray*} \left\vert\frac{1}{r}-\frac{140000}{4704}\right\vert & = & \l... ...ft\vert\frac{1}{g(x)}-\frac{1}{.056}\right\vert\hspace{8 mm}(*). \end{eqnarray*}$$

If only we knew...

$$\begin{eqnarray*} \left\vert\frac{1}{r}-\frac{140000}{4704}\right\vert<\eta & \R... ...& \left\vert r-\frac{4704}{140000}\right\vert<\frac{10}{140000}. \end{eqnarray*}$$

But from Exercise 26 we have

$$\begin{eqnarray*} \eta & = & \min\left(\frac{\epsilon a^2}{2},\frac{\vert a\vert}{2}\right) \end{eqnarray*}$$

$\begin{array}{cccccc} a & = & 140000/4704 & \epsilon & = & 10/140000\\ \epsilon a^2/2 & = &10\cdot 70000/4704^2 & a/2 & = & 70000/4704\vspace{2 mm} \end{array}$ $\begin{array}{ccccc} \eta & = & 700000/4704^2 & = & 1/.084+1/.056. \end{array}$

If only we knew...

$$\begin{array}{lcc} \vert 1/f(x)-1/.084\vert & < & \eta/2\\ \... ...+\frac{1}{g(x)}-\frac{140000}{4704}\right\vert < \eta \right..$$

And if only we knew...

$$\begin{eqnarray*} \vert f(x)-.084\vert < \beta_1 & \Rightarrow & \left\vert\fra... ...t\vert\frac{1}{g(x)}-\frac{1}{.056}\right\vert < \frac{\eta}{2}. \end{eqnarray*}$$

[ continued next page ] [ Exercise 1.36: Solution continued ]

But, again by Exercise 26, we have

$$\begin{eqnarray*} \beta_1=\min\left(\frac{\eta/2\hspace{2 mm}.084^2}{2}, \frac{... ...(\frac{\eta/2\hspace{2 mm}.056^2}{2}, \frac{.056}{2}\right)=... \end{eqnarray*}$$

``Frank, how close must $x$ be to 20 (precision $\delta_1$) so that $f(x)$ is tolerably close to .084 (tolerance=$\beta_1$)?''
``George, how close must $x$ be to 20 (precision $\delta_2$) so that $f(x)$ is tolerably close to .056 (tolerance=$\beta_2$)?''
Message to manufacturing: 2 wires, each with 20% copper at a precision of $\min(\delta_1,\delta_2)$.

`$\Box$