Module 4 - Limit as x Approaches a

Module 4 - Answers

Lesson 1

Answer 1

4.1.1 The smaller of the two choices, 0.13, will ensure that the output is within 0.1 of 2. If a point on the graph of the function has an x-coordinate that is within 0.13 of 3, then it will lie between the two horizontal lines and its y-coordinate will be within 0.1 of 2. This means that in order to achieve a y-tolerance of 0.1 you need to have an x-tolerance of 0.13 (or less.)

Answer 2

4.1.2

Graph
Set the viewing window to [2.98, 3.02] x [1.98, 2.02]
Find the intersection points by using the Intersect command

Use the intersection points to find the tolerance for x that produces a y-tolerance of 0.01.

Choose the smaller of the two tolerances. The y-tolerance is 0.01 when the x-tolerance is 0.0133.

Lesson 2

Answer 1

4.2.1 Find a positive number

so that 1.99 <

< 2.01 whenever 3 –

< x < 3 +

Answer 2

4.2.2 2.9867 < x < 3.01337

Answer 3

4.2.3

Solve 3 – 2.9867 and 3 + 3.01337 for to obtain 0.0133 and 0.01337.

The correct tolerance is the minimum of the two values, 0.0133.

Answer 4

4.2.4 For each positive

there is a positive

such that

5 –

< 5 +

whenever 10 –

< x < 10 +

means that 5 as x 10.

Lesson 3

Answer 1

4.3.1

= 1

Answer 2

4.3.2

Display the mode settings dialog box by pressing

The fourth item in the menu is the Angle mode.

Select Angle's setting by pressing three times

Open the Angle submenu by pressing
Highlight "Degree" by pressing

Select "Degree" by pressing
Save the changes and exit the Mode screen by press

The mode settings under the Edit Line should say DEG to indicate the calculator is currently in Degree mode.

The limit function should still be on the Edit Line.

Execute it again by pressing

Answer 3

4.3.3

represents the value that f(x) =

approaches as x approaches 0 from the left.

which is interpreted as

when x 0 from the left.

Answer 4

4.3.4

represents the value that approaches as x approaches 0 from the right.

which is interpreted as

when x 0 from the right.

Self Test

Answer 1

x should be within approximately 0.04881 of 1

Answer 2

Find a positive number so that 3.9 < x² + 3 < 4.1 whenever 1 – < x < 1 +

Because x is near 1, the positive solutions are what we need. Therefore,

1 – = 0.948683 and 1 + = 1.04881.

Solve for

Taking the minimum of the two values, x should be within approximately 0.04881 of 1.

Answer 3

Answer 4

Answers may vary but should include a graph of

that shows the vertical asymptotic behavior of the function at x = –2. The Trace cursor should have an x-coordinate slightly to the right of x = –2 and show a large negative y-coordinate.