12.1.2 An Introduction to L'Hôpital's Rule

Question 1

An Introduction to L’Hôpital’s Rule

Answer 1

• A limit of a function is called an indeterminate form when it produces a mathematically meaningless expression.
• L’Hôpital’s rule enables you to evaluate indeterminate forms quickly by using derivatives.

Question 2

note

Answer 2

• The limits of simple functions can be evaluated easily with simple algebra and direct numerical substitution.
• For complicated functions, it might not be easy to factor or simplify their respective limits, and in some cases, algebraic manipulation can be time-consuming and lead to errors.
• Suppose a function has a numerator and a denominator and its limit produces an indeterminate form. Then you can apply L’Hôpital’s rule.
• If the top and the bottom are differentiable, the limit of the function is equal to the limit of the derivative of the numerator divided by the derivative of the denominator.
• Instead of using algebra, L’Hôpital’s rule enables you to evaluate indeterminate forms quickly by using derivatives.
• If the quotient of the derivatives still yields an indeterminate form, L’Hôpital’s rule can be applied again.
• To avoid mistakes, make sure the limit produces an
  indeterminate form before using L’Hôpital’s rule.
• CAUTION: Notice that in the example above, the quotient rule is not applied when taking the derivatives. L’Hôpital’s rule doesn’t instruct you to take the derivative of the function. This makes L’Hôpital’s rule easier to use, since you do not have to remember the quotient rule.

Question 3

Evaluate limx→2 x^6−6/410x−5

Answer 3

0

Question 4

Evaluate limx→∞ x3+3x+1/4x2+2

Answer 4

∞

Question 5

Evaluate limx→0 x100+7x2/x50−4x2

Answer 5

−7/4

Question 6

Evaluate.

limx→9 x−9/x2−11x+18

Answer 6

1/7

Question 7

Evaluate limx→5x2−7x+10/x3−25x

Answer 7

0.06

Question 8

Evaluate limx→0 x6−x5+2x+4x5+3x−1

Answer 8

−4

Question 9

Evaluate limx→1 x3−2x+1/x4+3x−4

Answer 9

1/7

Question 10

Evaluate limx→2 x3−2x2/10x−20.

Answer 10

2/5

Question 11

Evaluate.

limx→1 x74−1/x148−1

Answer 11

1/2

Question 12

Evaluate.

limx→0 x2−2x2x−1

Answer 12

0

Question 13

Evaluate limx→2 x3−2x2+3x−6x2−4x+4

Answer 13

The limit does not exist

Question 14

Evaluate limx→3x3−9x2+27x−27x−3.

Answer 14

0