12.1.1 Indeterminate Forms

Question 1

Indeterminate Forms

Answer 1

• A limit of a function is called an indeterminate form when it produces a mathematically meaningless expression. Two types of indeterminate forms are 0/0 and ∞/∞.
• Some indeterminate forms can be solved by using algebraic tricks such as canceling or dividing by the highest power of x.

Question 2

note

Answer 2

• When taking limits, sometimes you will encounter expressions whose meanings can be interpreted in different ways. These limits are called indeterminate forms. 0/0 is one example.
• One camp says that the indeterminate form equals one because it is a number divided by itself.
• Another says that zero divided by anything is zero.
• A third says that any number divided by zero is infinity.
• Similar arguments hold for the form.
• When an indeterminate form arises, you will have to do more work.
• One algebraic trick involves factoring the numerator and the denominator.
• In this case, you can cancel the (x – 3) factors as long as you promise not to let x be equal to three.
• To evaluate this limit, look for the highest power.
• In this case, x 3 is the highest power, so divide the numerator and denominator by it. You are essentially multiplying by a form of one.
• Now all the terms have x in the denominator except one. Those terms will approach zero.
• The result is not an indeterminate form. It is negative infinity

Question 3

Evaluate limx→∞ 4x5+10x3+9x2+2x+12x5−3x4−9x+5

Answer 3

2

Question 4

Evaluate limx→3 x3−27x2−2x−3.

Answer 4

27/4

Question 5

Evaluate limx→∞ 8x8−x5+x2+11−2x8

Answer 5

-4

Question 6

Evaluate limx→100100x−x2x3−100x2.

Answer 6

-1/100

Question 7

Evaluate limx→2 x3−4x3x−6.

Answer 7

8/3

Question 8

Evaluate limx→1 x12−2x11+x10x3−x2−x+1

Answer 8

1/2

Question 9

Evaluate limx→∞ 2x3−33x3−4x2+1.

Answer 9

2/3

Question 10

Evaluate limx→3 2x−6x2−4x+3.

Answer 10

1

Question 11

Evaluate limx→∞ 2x2000−5x+4002x2001+2001

Answer 11

0

Question 12

Evaluate limx→2 x3−2x210x−20

Answer 12

2/5

Question 13

Evaluate limx→0 x3−3x211x5−4x2.

Answer 13

3/4

Question 14

Evaluate limx→1 x2+2x−3x−1

Answer 14

4