Chapter 4 | Applications of Derivatives
455
Figure 4.71 If lim x → a g ( x ), then the ratio f ( x )/ g ( x ) is approximately equal to the ratio of their linear approximations near a . f ( x ) = lim x → a
Since f is differentiable at a , then f is continuous at a , and therefore f ( a ) = lim x → a
f ( x ) =0. Similarly,
g ( a ) = lim x → a g ′( a ) = lim x → a
g ( x ) =0. If we also assume that f ′ and g ′ are continuous at x = a , then f ′( a ) = lim x → a
f ′( x ) and
g ′( x ). Using these ideas, we conclude that
lim x → a f ′( x ) g ′( x ) . Note that the assumption that f ′ and g ′ are continuous at a and g ′( a ) ≠0 can be loosened. We state L’Hôpital’s rule formally for the indeterminate form 0 0 . Also note that the notation 0 0 does not mean we are actually dividing zero by zero. Rather, we are using the notation 0 0 to represent a quotient of limits, each of which is zero. f ( x ) g ( x ) = lim x → a f ′( x )( x − a ) g ′( x )( x − a ) = lim x → a Theorem 4.12: L’Hôpital’s Rule (0/0 Case) Suppose f and g are differentiable functions over an open interval containing a , except possibly at a . If lim x → a f ( x ) =0 and lim x → a g ( x ) =0, then lim x → a f ( x ) g ( x ) = lim x → a f ′( x ) g ′( x ) , assuming the limit on the right exists or is ∞ or −∞. This result also holds if we are considering one-sided limits, or if a =∞and −∞. Proof We provide a proof of this theorem in the special case when f , g , f ′, and g ′ are all continuous over an open interval containing a . In that case, since lim x → a f ( x ) =0= lim x → a g ( x ) and f and g are continuous at a , it follows that f ( a ) =0= g ( a ). Therefore,
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