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l’Hôpital’s Rule Help

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By — McGraw-Hill Professional
Updated on Apr 25, 2014

Introduction to l’Hôpital’s Rule

Consider the limit

Indeterminate Forms 5.1 l’Hôpital’s Rule

If lim xc f ( x ) exists and lim xc g ( x ) exists and is not zero then the limit (*) is straightforward to evaluate. However, as we saw in Theorem 2.3, when lim xc g ( x ) = 0 then the situation is more complicated (especially when lim xc f ( x ) = 0 as well).

For example, if f ( x ) = sin x and g ( x ) = x then the limit of the quotient as x → 0 exists and equals 1. However if f ( x ) = x and g ( x ) = x 2 then the limit of the quotient as x → 0 does not exist.

In this section we learn a rule for evaluating indeterminate forms of the type (*) when either lim xc f ( x ) = lim xc g ( x ) = 0 or lim xc f ( x ) = lim xc g ( x ) = ∞. Such limits, or “forms,” are considered indeterminate because the limit of the quotient might actually exist and be finite or might not exist; one cannot analyze such a form by elementary means. 

L’Hôpital’s Rule

Theorem 5.1 (l’Hôpital’s Rule)

Let f ( x ) and g ( x ) be differentiable functions on ( a, c ) ∪ ( c, b ). If

Indeterminate Forms 5.1 l’Hôpital’s Rule

then

Indeterminate Forms 5.1 l’Hôpital’s Rule

provided this last limit exists as a finite or infinite limit.

Let us learn how to use this new result.

Example 1

Evaluate

Indeterminate Forms 5.1 l’Hôpital’s Rule

Solution 1

We first notice that both the numerator and denominator have limit zero as x tends to 1. Thus the quotient is indeterminate at 1 and of the form 0/0. l’Hôpital’s Rule therefore applies and the limit equals

Indeterminate Forms 5.1 l’Hôpital’s Rule

provided this last limit exists. The last limit is

Indeterminate Forms 5.1 l’Hôpital’s Rule

Therefore we see that

Indeterminate Forms 5.1 l’Hôpital’s Rule

You Try It : Apply l’Hôpital’s Rule to the limit lim x →2 sin( π x )/( x 2 − 4).

You Try It : Use l’Hôpital’s Rule to evaluate lim h →0 (sin h/h ) and lim h →0 (cos h − 1/ h ). These limits are important in the theory of calculus.

Example 2

Evaluate the limit

Indeterminate Forms 5.1 l’Hôpital’s Rule

Solution 2

As x → 0 both numerator and denominator tend to zero, so the quotient is indeterminate at 0 of the form 0/0. Thus l’Hôpital’s Rule applies. Our limit equals

Indeterminate Forms 5.1 l’Hôpital’s Rule

provided that this last limit exists. It equals

Indeterminate Forms 5.1 l’Hôpital’s Rule

This is another indeterminate form. So we must again apply l’Hôpital’s Rule. The result is

Indeterminate Forms 5.1 l’Hôpital’s Rule

This is again indeterminate; another application of l’Hôpital’s Rule gives us finally

Indeterminate Forms 5.1 l’Hôpital’s Rule

We conclude that the original limit equals 6.

You Try It : Apply l’Hôpital’s Rule to the limit lim x →0 x /[1 / ln | x |].

Indeterminate Forms Involving Infinity

Indeterminate Forms Involving ∞ We handle indeterminate forms involving infinity as follows: Let f ( x ) and g ( x ) be differentiable functions on ( a, c ) ∪ ( c, b ). If

Indeterminate Forms 5.1 l’Hôpital’s Rule

both exist and equal +∞ or −∞ (they may have the same sign or different signs) then

Indeterminate Forms 5.1 l’Hôpital’s Rule

provided this last limit exists either as a finite or infinite limit.

Let us look at some examples.

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