l’Hôpital’s Rule Help
Introduction to l’Hôpital’s Rule
Consider the limit
If lim x → c f ( x ) exists and lim x → c g ( x ) exists and is not zero then the limit (*) is straightforward to evaluate. However, as we saw in Theorem 2.3, when lim x → c g ( x ) = 0 then the situation is more complicated (especially when lim x → c 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 x → c f ( x ) = lim x → c g ( x ) = 0 or lim x → c f ( x ) = lim x → c 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.
Theorem 5.1 (l’Hôpital’s Rule)
Let f ( x ) and g ( x ) be differentiable functions on ( a, c ) ∪ ( c, b ). If
provided this last limit exists as a finite or infinite limit.
Let us learn how to use this new result.
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
provided this last limit exists. The last limit is
Therefore we see that
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.
Evaluate the limit
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
provided that this last limit exists. It equals
This is another indeterminate form. So we must again apply l’Hôpital’s Rule. The result is
This is again indeterminate; another application of l’Hôpital’s Rule gives us finally
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
both exist and equal +∞ or −∞ (they may have the same sign or different signs) then
provided this last limit exists either as a finite or infinite limit.
Let us look at some examples.
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