MATH 155 SUPPLEMENTAL NOTES 6

INDETERMINATE FORMS AND L' Hôpital's Rule

INDETERMINATE FORMS

Consider the following limit.

First of all, 0/0 ¹ 0. In fact, 0/0 is a meaningless expression known as an indeterminate form. There are seven known indeterminate forms. They are 0/0, ¥ /¥ , 0 * ¥ , ¥ - ¥ , 1^¥ , 0 0, ¥ 0. In the following notes, I will discuss how to handle each form. A method that is used for most of these cases is called L' Hôpital's Rule.

L' Hôpital's Rule

L' Hôpital's Rule

Suppose that f(a) = g(a) = 0 and that f and g are differentiable on an open interval I containing a. Suppose also that g' (x) ¹ 0 on I if x ¹ a. Then

if the limit on the right exists (or is ± ¥ ).

Now, I will apply L' Hôpital's Rule to these different indeterminate forms. I will break these examples into three different cases.

CASE 1: 0 / 0 and ¥ / ¥

EXAMPLE 1:

SOLUTION:

First of all, determine if the limit results in an indeterminate form. As x ® 0, (sin x)/x ® 0/0, so it is an indeterminate form. Therefore, we can apply L' Hôpital's Rule.

After applying L' Hôpital's Rule check the following limit to determine if it is an indeterminate form or a regular limit. If it is an indeterminate form, then apply L' Hôpital's Rule again. If not, then evaluate the limit.

EXAMPLE 2:

SOLUTION:

Looking at this limit, I notice that as x ® 2, the limit approaches 0/0, so it is an indeterminate form. Therefore, I will apply L' Hôpital's Rule to determine this limit.

EXAMPLE 3:

SOLUTION:

As x ® ¥ , the limit approaches ¥ / ¥ , therefore its result is an indeterminate form. We will now apply L' Hôpital's Rule to this limit.

Notice that each following limit is the indeterminate form ¥ / ¥ , so I keep applying L' Hôpital's Rule until I reach a limit that I can find by methods of calculus I.

EXAMPLE 4:

SOLUTION:

I will first simplify the expression that is inside the limit using the change of base formula for logarithms. If you do not remember this formula, then review supplemental notes 4.

As x ® ¥ , the limit approaches ¥ / ¥ . Therefore it approaches an indeterminate form, and L' Hôpital's Rule can safely be applied.

CASE 2: 0 * ¥ and ¥ - ¥

NOTE: 0 * ¥ ¹ 0 since ¥ is not a fixed number

NOTE: ¥ - ¥ ¹ 0, since one ¥ is not the same as another ¥ .

To evaluate limits that result in these two indeterminate forms, use algebra to get them into the indeterminate forms 0/0 or ¥ / ¥ . Then apply L' Hôpital's Rule.

EXAMPLE 5:

SOLUTION:

This limit evaluates to 0 * ¥ , so we must convert the x cot x to a form that will either get us 0/0 or ¥ / ¥ . Recall that cot x = (cos x)/(sin x), therefore the limit becomes the following.

Now the limit evaluates to the indeterminate form of 0/0.

EXAMPLE 6:

SOLUTION:

This limit evaluates to the form of 0 * ¥ , so we will have to convert the expression to one that will result in a limit of either 0/0 or ¥ /¥ . To do this, I will use a dummy variable. Let u = 1/x, then x = 1/u. As x ® ¥ , then u ® 0.

EXAMPLE 7:

SOLUTION:

This limit evaluates to the indeterminate form ¥ - ¥ . To convert this limit into one that is either 0/0 or ¥ / ¥ , I will find a common denominator for the expression.

Now this limit evaluates to be 0/0, so let us apply L' Hôpital's Rule.

EXAMPLE 8:

SOLUTION:

This limit evaluates to ¥ - ¥ , so to convert this into expression that results in a limit of 0/0 or ¥ /¥ , I will rationalize the numerator of this fraction.

After rationalizing the numerator of the expression, the limit becomes a straightforward. Remember as the denominator gets larger, 1 divided by an extremely large number approaches 0.

 CASE 3: 1^ ¥, 0 0, ¥ 0

NOTE: 1^ ¥ ¹ 1 because infinity is not a finite number.

NOTE: 0 0 ¹ 1 because 0/0 = 0 1 - 1 = 0 0.

NOTE: ¥ 0 ¹ 1 because infinity is not a finite number.

To find limits of functions that result in these indeterminate forms, we will take logarithms first, then apply the following fact.

FACT:

EXAMPLE 9:

SOLUTION:

This functions limit is of the form 0 0.

f (x) = x x ® ln f (x) = x ln x

The first limit evaluates to 0 * ¥. The second limit evaluates to ¥ /¥. Now, to complete this problem, we will now apply the above fact.

EXAMPLE 10:

SOLUTION:

This functions limit is of the form ¥ ^ 0.

Therefore,

EXAMPLE 11:

 

SOLUTION:

This functions limit is of the form 1^ ¥ .

Therefore,

Work through all three cases carefully noting the way each indeterminate form is handled. If you have any questions or problems with any of these examples, feel free to contact me.

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