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Chapter 4 | Applications of Derivatives
4.8 | L’Hôpital’s Rule
Learning Objectives
4.8.1 Recognize when to apply L’Hôpital’s rule. 4.8.2 Identify indeterminate forms produced by quotients, products, subtractions, and powers, and apply L’Hôpital’s rule in each case. 4.8.3 Describe the relative growth rates of functions. In this section, we examine a powerful tool for evaluating limits. This tool, known as L’Hôpital’s rule , uses derivatives to calculate limits. With this rule, we will be able to evaluate many limits we have not yet been able to determine. Instead of relying on numerical evidence to conjecture that a limit exists, we will be able to show definitively that a limit exists and to determine its exact value. Applying L’Hôpital’s Rule L’Hôpital’s rule can be used to evaluate limits involving the quotient of two functions. Consider lim x → a f ( x ) g ( x ) . If lim x → a f ( x ) = L 1 and lim x → a g ( x ) = L 2 ≠0, then lim x → a f ( x ) g ( x ) = .
L 1 L 2
g ( x ) =0? We call this one of the indeterminate forms , of type 0 0 .
f ( x ) =0 and lim x → a
However, what happens if lim x → a
This is considered an indeterminate form because we cannot determine the exact behavior of f ( x ) g ( x ) further analysis. We have seen examples of this earlier in the text. For example, consider lim x →2 x 2 −4 x −2 and lim x →0 sin x x . For the first of these examples, we can evaluate the limit by factoring the numerator and writing lim x →2 x 2 −4 x −2 = lim x →2 ( x +2)( x −2) x −2 = lim x →2 ( x +2)=2+2=4. For lim x →0 sin x x we were able to show, using a geometric argument, that lim x →0 sin x x =1.
as x → a without
Here we use a different technique for evaluating limits such as these. Not only does this technique provide an easier way to evaluate these limits, but also, and more important, it provides us with a way to evaluate many other limits that we could not calculate previously. The idea behind L’Hôpital’s rule can be explained using local linear approximations. Consider two differentiable functions f and g such that lim x → a f ( x ) =0= lim x → a g ( x ) and such that g ′( a ) ≠0 For x near a , we can write f ( x ) ≈ f ( a )+ f ′( a )( x − a ) and g ( x ) ≈ g ( a )+ g ′( a )( x − a ). Therefore, f ( x ) g ( x ) ≈ f ( a )+ f ′( a )( x − a ) g ( a )+ g ′( a )( x − a ) .
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