Example 2 (page 74 in Stewart)
The example uses a calculator to look for evidence about the limit.
key to call the function editor,
and enter the function.
________________________________
| |
| Y1=(
(X^2+9)-3)/X^2 |
| Y2= |
| Y3= |
| Y4= |
| Y5= |
| |
| |
| |
| |
|________________________________|
.
to enter a " : ".
key to do the calculation.
Your screen should look like this:
________________________________
| |
| 1->X:Y1 |
| .1622776602 |
| |
| |
| |
| |
| |
| |
| |
|________________________________|
to recall your previous entry.
________________________________
| |
| 1->X:Y1 |
| .1622776602 |
| .01->X:Y1 |
| .166666204 |
| .00001->X:Y1 |
| .167 |
| .0000001->X:Y1 |
| 0 |
| |
|________________________________|
Obviously there are problems with the calculations.
You should read Stewart's comments on the top of page 75,
where he points out that the problem is that
(x^2+9)
is very close to 3 when x is small.
The program in the calculator
is smart enough to recognize the difficulty,
so it gives a smaller and smaller number
of significant digits.
Example 6 (page 88 in Stewart)
After the problems calculating the above limit, it is interesting to look at this example. Stewart shows that we can use algebra to change the form of the limit.
=
________________________________
| |
| Y1=(
(X^2+9)-3)/X^2 |
| Y2=1/(
(X^2+9)+3) |
| Y3= |
| Y4= |
| Y5= |
| |
| |
| |
| |
|________________________________|
________________________________
| |
| 1->X:Y2 |
| .1622776602 |
| .01->X:Y2 |
| .1666662037 |
| .00001->X:Y2 |
| .1666666666 |
| .0000001->X:Y2 |
| .1666666667 |
| |
|________________________________|