ACOW LIMITS AND CONTINUITY MODULE
Updated 5/28/2016
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Activity A3 – Finding Limits Numerically
In activities A1 and A2 you used the limit applet to graph a function and then determine
the limit of a function at a particular value of x. Now we will use the limit applet(RS1) to
investigate the limit of functions numerically (using a table).
Suppose we want to find
lim  2 x 2  3 .
x 2
To do this, we need to find lim  2 x  3 and lim  2 x 2  3 . If the left-hand limit is the
2
x 2
x 2
same as the right-hand limit, then the limit as x goes to 2 will exist. In the limit applet,
scroll down to the bottom of the applet and enter the function into the f(x) cell (see
below).
To find the lim  2 x 2  3 , we need to fill in the x cells with values of x that lie on the
x 2
right side of 2, but are real close to 2. The values in the cells below are good values to
choose.
Once the table is filled in click on Fill Table. The f(x) values will appear below the
correcsponding x values (see below).
Notice the function’s values seem to approach 11 as x approaches 2 on its right side.
Thus, lim  2 x 2  3  11 .
x 2
ACOW LIMITS AND CONTINUITY MODULE
Updated 5/28/2016
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Now create a chart to find lim  2 x 2  3 (see below).
x 2
Notice the function’s values seem to approach 11 as x approaches 2 on its left side. Thus,
lim  2 x 2  3  11 and consequently, lim  2 x 2  3  11 . You can confirm this
x 2
x 2
graphically by using the upper portion of the applet as we did in activities A1 and A2.
1. Let f ( x)  2 x  3 and use the table at the bottom of the limit applet(RS1) to
complete the tables and answer the questions that follow.
a)
10.7
10.8
10.9
10.99
10.999
x
f (x)
b) lim
x 11


2x  3 
c)
11.3
x
f (x)
d) lim
x 11
e) lim
x 11


11.2
11.1
11.01
11.001

2x  3 

2x  3 
x 1
and use the table at the bottom of the limit applet(RS1) to complete
x3
the tables and answer the questions that follow.
a)
2.7
2.8
2.9
2.99
2.999
x
f (x)
2. Let f ( x) 
b) lim
x 3
x 1
x3
c)
x
f (x)
3.3
3.2
3.1
3.01
3.001
ACOW LIMITS AND CONTINUITY MODULE
Updated 5/28/2016
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x 1
x 3 x  3
x 1
e) lim
x 3 x  3
d) lim
1
and use the table at the bottom of the limit applet(RS1) to
( x  2) 2
complete the tables and answer the questions that follow.
a)
1.7
1.8
1.9
1.99
1.999
x
f (x)
3. Let f ( x) 
b) lim
x 2
1
( x  2) 2
c)
x
f (x)
2.3
2.2
2.1
2.01
2.001
1
2
x  2 ( x  2)
1
e) lim
x  2 ( x  2) 2
d) lim
We can also use the table at the bottom of the limit applet to find limits of piecewise
functions like one shown below.
Since x < 1 represent x values on the left side of 1, we will use the function 2x + 5 to find
the left hand limit (see below).
ACOW LIMITS AND CONTINUITY MODULE
Updated 5/28/2016
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The function’s values seem to approach 7, thus lim f ( x)  7 .
x 1
Since x  1 represent x values on the right side of 1, we will use the function x 2 to find
the right hand limit (see below).
The function’s values seem to approach 1, thus lim f ( x)  1 . Since lim f ( x)  lim f ( x) ,
x 1
x 1
lim f ( x)  DNE .
x 1
Use the table at the bottom of the limit applet(RS1) to find the following
 x  7 x  3
4. lim  2
x 3 x  19
x  3

x  3 x  1

5. lim  5
x 1
x 1
 x2
x 1

x0
 3x

6. lim  x  1 0  x  2
x 0
2 x  1
x2

x0
 3x

7. lim  x  1 0  x  2
x 2
2 x  1
x2

x 1