L’Hôpital’s Rule: Learn It 3

Other Indeterminate Forms

L’Hôpital’s rule is very useful for evaluating limits involving the indeterminate forms 00 and . However, we can also use L’Hôpital’s rule to help evaluate limits involving other indeterminate forms that arise when evaluating limits.

The expressions 0, , 1, 0, and 00 are all considered indeterminate forms. These expressions are not real numbers. Rather, they represent forms that arise when trying to evaluate certain limits.

Next we realize why these are indeterminate forms and then understand how to use L’Hôpital’s rule in these cases. The key idea is that we must rewrite the indeterminate forms in such a way that we arrive at the indeterminate form 00 or .

Indeterminate Form of Type 0

Suppose we want to evaluate limxa(f(x)g(x)), where f(x)0 and g(x) (or ) as xa.

Since one term in the product is approaching zero but the other term is becoming arbitrarily large (in magnitude), anything can happen to the product. We use the notation 0 to denote the form that arises in this situation.

The expression 0 is considered indeterminate because we cannot determine without further analysis the exact behavior of the product f(x)g(x) as xa. For example, let n be a positive integer and consider

f(x)=1(xn+1) and g(x)=3x2.

As x, f(x)0 and g(x).

However, the limit as x of f(x)g(x)=3x2(xn+1) varies, depending on n. If n=2, then limxf(x)g(x)=3. If n=1, then limxf(x)g(x)=. If n=3, then limxf(x)g(x)=0.

Here we consider another limit involving the indeterminate form 0 and show how to rewrite the function as a quotient to use L’Hôpital’s rule.

Evaluate limx0+xlnx.

First, rewrite the function xlnx as a quotient to apply L’Hôpital’s rule. If we write

xlnx=lnx1/x,

we see that lnx as x0+ and 1x as x0+. Therefore, we can apply L’Hôpital’s rule and obtain

limx0+lnx1/x=limx0+ddx(lnx)ddx(1/x)=limx0+1/x1/x2=limx0+(x)=0.

We conclude that

limx0+xlnx=0.
The function y = x ln(x) is graphed for values x ≥ 0. At x = 0, the value of the function is 0.
Figure 2. Finding the limit at x=0 of the function f(x)=xlnx.

Evaluate limx0xcotx.

Indeterminate Form of Type

Another type of indeterminate form is . Consider the following example:

Let n be a positive integer and let f(x)=3xn and g(x)=3x2+5.

As x, f(x) and g(x). We are interested in limx(f(x)g(x)).

Depending on whether f(x) grows faster, g(x) grows faster, or they grow at the same rate, as we see next, anything can happen in this limit. Since f(x) and g(x), we write to denote the form of this limit.

As with our other indeterminate forms, has no meaning on its own and we must do more analysis to determine the value of the limit.

Suppose the exponent n in the function f(x)=3xn is n=3, then

limx(f(x)g(x))=limx(3x33x25)=.

On the other hand, if n=2, then

limx(f(x)g(x))=limx(3x23x25)=5.

However, if n=1, then

limx(f(x)g(x))=limx(3x3x25)=.

Therefore, the limit cannot be determined by considering only

Next we see how to rewrite an expression involving the indeterminate form as a fraction to apply L’Hôpital’s rule.

Evaluate limx0+(1x21tanx).

By combining the fractions, we can write the function as a quotient. Since the least common denominator is x2tanx, we have

1x21tanx=(tanx)x2x2tanx

As x0+, the numerator tanxx20 and the denominator x2tanx0. Therefore, we can apply L’Hôpital’s rule. Taking the derivatives of the numerator and the denominator, we have

limx0+(tanx)x2x2tanx=limx0+(sec2x)2xx2sec2x+2xtanx

As x0+, (sec2x)2x1 and x2sec2x+2xtanx0. Since the denominator is positive as x approaches zero from the right, we conclude that

limx0+(sec2x)2xx2sec2x+2xtanx=

Therefore,

limx0+(1x21tanx)= 

Watch the following video to see the worked solution to this example.

Evaluate limx0+(1x1sinx).

Indeterminate Form of Limits Involving Exponents

Another type of indeterminate form that arises when evaluating limits involves exponents. The expressions 00, 0, and 1 are all indeterminate forms.

On their own, these expressions are meaningless because we cannot actually evaluate these expressions as we would evaluate an expression involving real numbers. Rather, these expressions represent forms that arise when finding limits.

Since L’Hôpital’s rule applies to quotients, we use the natural logarithm function and its properties to reduce a problem evaluating a limit involving exponents to a related problem involving a limit of a quotient.

Suppose we want to evaluate limxaf(x)g(x) and we arrive at the indeterminate form 0. (The indeterminate forms 00 and 1 can be handled similarly.) 

Let

y=f(x)g(x)

Then,

lny=ln(f(x)g(x))=g(x)ln(f(x))

Therefore,

limxa[lny]=limxa[g(x)ln(f(x))]

Since limxaf(x)=, we know that limxaln(f(x))=. Therefore, limxag(x)ln(f(x)) is of the indeterminate form 0, and we can use the techniques discussed earlier to rewrite the expression g(x)ln(f(x)) in a form so that we can apply L’Hôpital’s rule.

Suppose limxag(x)ln(f(x))=L, where L may be or . Then

limxalny=L

Since the natural logarithm function is continuous, we conclude that

ln(limxay)=L

which gives us

limxay=limxaf(x)g(x)=eL

Evaluate limxx1x

Evaluate limx0+xsinx