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Calculus- Early Transcendentals, 2021a

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192 Applications of Derivatives<br />

Definition 5.33: Limits of the Indeterminate Forms 0 0 and ∞ ∞<br />

f (x)<br />

A limit of a quotient lim<br />

x→a g(x) is said to be an indeterminate form of the type 0 0 if<br />

both f (x) → 0 and g(x) → 0 as x → a.<br />

Likewise, it is said to be an indeterminate form of the type ∞ ∞ if<br />

both f (x) →±∞ and g(x) →±∞ as x → a.<br />

Note that the two ± signs are independent of each other.<br />

Theorem 5.34: L’Hôpital’s Rule<br />

f (x)<br />

For a limit lim<br />

x→a g(x) of the indeterminate form 0 0 or ∞ ∞ ,<br />

f<br />

if lim<br />

′ (x)<br />

x→a g ′ (x)<br />

exists or equals ∞ or −∞.<br />

f (x)<br />

lim<br />

x→a g(x) = lim f ′ (x)<br />

x→a g ′ (x)<br />

This theorem is somewhat difficult to prove, in part because it incorporates so many different possibilities,<br />

so we will not prove it here.<br />

We should also note that there may be instances where we would need to apply L’Hôpital’s Rule<br />

f<br />

multiple times, but we must confirm that lim ′ (x)<br />

x→a g ′ (x)<br />

is still indeterminate before we attempt to apply<br />

L’Hôpital’s Rule again. Finally, we want to mention that L’Hôpital’s rule is also valid for one-sided limits<br />

and limits at infinity.<br />

Example 5.35: L’Hôpital’s Rule<br />

x 2 − π 2<br />

Compute lim<br />

x→π sinx .<br />

Solution. We use L’Hôpital’s Rule: Since the numerator and denominator both approach zero,<br />

x 2 − π 2 2x<br />

lim = lim<br />

x→π sinx x→π cosx ,<br />

provided the latter exists. But in fact this is an easy limit, since the denominator now approaches −1, so<br />

x 2 − π 2<br />

lim = 2π<br />

x→π sinx −1 = −2π.<br />

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