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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.
Solution. Since csc x ln(x + 1) assumes the form ∞∙0 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
Solution. This is of type ∞ - ∞.
using L”Hopital’s Rule
Indeterminate forms 1 ∞, 00, ∞0. If
takes on one of the indeterminate forms 1 ∞, 00, ∞0 , 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.]
then the limit desired is
We can treat type
in a similar way since
And again the method applies to type
To find lim fg which assumes an indeterminate form of type 1 ∞, 00, ∞0 apply L’Hopital’s rule to
If L’Hopital’s Rule applied to one of these yields b for a limit, then lim fg = eb.
Solution. This is of type 1∞. Applying L’Hopital’s Rule to
so the required limit is
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