Using the TI-85 graphing calculator, find a d which corresponds to e = 0.1 in the
definition of limits to obtain evidence that
Let f(x) = x3 + 5x + 1.
We want to find the d which corresponds to e = 0.1.
The strategy is to determine what values of x
makes the function y = |f(x) - 19| - 0.1
less than 0 near a = 2.
[Press here to see animation again!]
- If you are not familiar with the graphing of functions on the TI-85,
then first read the
Little's Basic Guide to the TI-85.
- Press the GRAPH key and then pick y(x)=
by pressing the F1 key.
- If necessary, keep pressing F4 key until only
y1= appears on the screen.
- Type in the function
abs(x^3 + 5 x + 1 - 19) - 0.1
and press the ENTER key. You can get the function
abs by pressing 2nd CATALOGUE and the
- Press EXIT and the F2 keys
to pick RANGE from the menu.
- Let us begin with the viewing window [ 1, 3] × [-1, 1]:
- Press the F5 key to graph the function:
- To get a clearer picture of what is happening, we shall zoom in. Press
F3 key to pick ZOOM from the menu.
- Now press F2 key to pick ZIN, the command
for zooming in.
- Press the ENTER key several times to get a good
- We shall now find the two roots of our functions which are shown in the
above graph. Press the EXIT key twice folowed by the
MORE key so that MATH appears on the menu.
- Now press F1 key to pick MATH.
- Press F3 key to pick ROOT.
- Use the arrow keys to move the blinking cursor close to one of the
roots and press the ENTER key. The following
shows the left-hand root:
which is 1.9941053957... .
- Repeat the preceding step to find the other root:
which is 2.005870179... .
- We see that the interval [1.9945, 2.0055] lies between the two roots
and, hence, if we choose d = 0.0055 then we have that 0 < |x - 2| < 0.0055 implies that |f(x) - 19| < 0.1.
- We can visualize this by graphing the function f(x) = x3 + 5 x + 1 using the window [1.9945, 2.0055] × [18.9, 19.1]:
Acknowledgment: The first place which the author saw this way to approach
this problem is in Jerry Johnson and Benny Evans' Discovering Calculus with