limits and the calculator
1. Let f(x) =
1  cos x
x2
. For each of the following values of x, use your calculator to compute f(x).
Give answers accurate to 5 places past the decimal.
x
10-1
10-2
10-2
10-3
10-4
10-5
10-6
10-7
10-8
10-9
10-10
10-11
f(x)
x
f(x)
On the basis of these numbers, what do you think is the value of lim lim
1  cos x
x 0
2. Show that the expression
1 cos x
x2
1  cos x
x2
is algebraically equivalent to
?
(sin x)2
. Hint: multiply
x2 (1  cosx)
top and bottom by (1 + cosx) and apply your favorite trig identity.
Using this equivalent algebraic form f(x) =
x
x2
(sin x ) 2
x 2 (1  cos x )
, redo the above table.
10-1
10-2
10-2
10-3
10-4
10-5
10-6
10-7
10-8
10-9
10-10
10-11
f(x)
x
f(x)
On the basis of these numbers, what do you think is the value of lim
x 0
1  cos x
x2
? Which is the
correct answer? What's going on? What's the moral of the story?
3a. Make sure your calculator is in the radian mode and then enter in the function
y = sin(1/x). Graph this function using the zoom6 screen. (Zoom6 gives a standard 10x10 viewing
1
screen.) Make a guess as to what you think is the value of lim sin   .
x 0
x
3b. Zoom in on your graph by entering Zoom2. Look at the graph and then zoom in again. Repeat
this process several times. Then make another guess as to what you
1
think is the value of lim sin   . What's going on? What's the moral of the story?
x 0
x