INDETERMINATE FORMS, L’HOPITAL’S RULE

The indeterminate forms 0/0 and . If two functions f(x) and g(x) both become zero at x = a, or if both become infinite as x approaches a, their quotient f(x)/g(x) is without meaning when x = a. In the first case the fraction is said to assume the indeterminate form 0/0 when x = a; in the second case it is said to assume the indeterminate form when x approaches a. An example that occurs frequently is the indeterminate form 0/0 assumed by Δy/Δx when Δx approaches zero in the definition of the derivative.

The indeterminate forms 0/0 and may arise when x becomes infinite as well as for some finite value of x. Thus if f(x) or g(x) both become zero or become infinite as x becomes infinite, the fraction f(x)/g(x) assumes one of the indeterminate forms 0/0 or .                                   

Def. Indeterminate form. An expression of type

Such expressions are undefined. They arise in various ways, especially when replacing different members of composite functions with their limits.

L’Hopital’s Rule. If the fraction f(x)/g(x) assumes one of the indeterminate forms 0/0 or when x = a, then

A similar statement holds if the indeterminate form arises when x becomes infinite instead of when x = a.

If f '(x)/g'(x) also assumes an indeterminate form of the type 0/0 or , the rule can be applied likewise to it. The process can, in fact, be repeated as many times as necessary to yield a fraction f⁽ⁿ⁾(a) / g⁽ⁿ⁾(a) which is not indeterminate.

Proof. For the case of indeterminates of form 0/0, L’Hopital’s rule follows directly from Cauchy’s Generalized Mean Value Theorem which states that if f(x) and g(x) are differentiable functions over an open interval (a, b) then

Write x for b in 1) and it becomes

Then, since f(a) = g(a) = 0, equation 2) becomes

Now taking limits of both sides and noting that as x approaches a, x₁ also approaches a

L’Hopital’s rule can be proved in general under the assumption that there exists a neighborhood U of a in which f and g are both differentiable, except possibly at a, and there is no point of U at which f ' and g' are both simultaneously zero.

Example 1. Evaluate

Solution. When θ = 0, both numerator and denominator are zero. Thus it is of type 0/0 and the rule applies.

Example 2. Evaluate

Solution. When x = 2, both numerator and denominator are zero, it is of type 0/0, and the rule applies.

The indeterminate form 0 • . If f(x)g(x) assumes the indeterminate form 0 • when x has a finite value a, or when x becomes positively or negatively infinite, the product may be converted to one of the indeterminate forms 0/0 or by use of the identities

and hence may be investigated by one of the foregoing rules.

Example. Evaluate

Solution. Since csc x ln(x + 1) assumes the form for x = 0, we write

This assumes the form 0/0 for x = 0. Hence by L’Hopital’s rule

The indeterminate form . An indeterminate form of type lim (f - g) = can be transformed into a type 0/0 using the algebraic identity

Example. Evaluate

Solution. This is of type .

using L”Hopital’s Rule

Indeterminate forms . If

takes on one of the indeterminate forms , then taking the logarithm will convert it into a type 0 • . Then from a type 0 • we can convert it into a type 0/0 or . Let us illustrate with an example. Suppose

Taking the logarithm we have

Now we convert it to a 0/0 type by using

and then apply L’Hopital’s Rule. [Or, alternatively , we can write

and apply the rule if this form is easier to handle as far as the differentiation is concerned.] 

If

then the limit desired is

We can treat type

in a similar way since

And again the method applies to type

since

Summary.

To find lim f^g which assumes an indeterminate form of type apply L’Hopital’s rule to

If L’Hopital’s Rule applied to one of these yields b for a limit, then lim f^g = e^b.

Example. Evaluate

Solution. This is of type Appling L’Hopital’s Rule to

gives

so the required limit is