When finding limits by hand, we often need to make an algebraic change that lets us see what the limit will be. Even when using a calculator to approximate a limit, the same algebraic changes may be necessary.

**Example 2** (page 74 in Stewart)

The example uses a calculator to look for evidence about the limit.

On your calculator, the entry for Y

________________________________ | | | Y_{1}=((X^2+9)-3)/X^2 | | Y_{2}= | | Y_{3}= | | Y_{4}= | | Y_{5}= | | | | | | | | | |________________________________|

This is done by entering 1 followed by the keystrokes .

Use the keystrokes to enter a " : ".

Now use the Y-variables menu to enter y

(See the "Function Evaluation" tutorial if you are unsure how to call the Y-variables.)

Now press the key to do the calculation. Your screen should look like this:

________________________________ | | | 1->X:Y_{1}| | .1622776602 | | | | | | | | | | | | | | | |________________________________|

Then you can use the cursor keys to change the input for x, to do the remaining parts of the question.

Try these values for x: 1, .01, .00001, and .0000001.

Your screen should look like this:

________________________________ | | | 1->X:YObviously 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._{1}| | .1622776602 | | .01->X:Y_{1}| | .166666204 | | .00001->X:Y_{1}| | .167 | | .0000001->X:Y_{1}| | 0 | | | |________________________________|

**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.

=

Although Stewart computes the limit algebraically, it is interesting to compute the values of the new function for exactly the same values of x that we used above. You will see that changing the algebraic form of the function avoids the problems with calculation.________________________________ | | | Y_{1}=((X^2+9)-3)/X^2 | | Y_{2}=1/((X^2+9)+3) | | Y_{3}= | | Y_{4}= | | Y_{5}= | | | | | | | | | |________________________________|

________________________________ | | | 1->X:Y_{2}| | .1622776602 | | .01->X:Y_{2}| | .1666662037 | | .00001->X:Y_{2}| | .1666666666 | | .0000001->X:Y_{2}| | .1666666667 | | | |________________________________|