Calculus 2nd Edition Rogawski

SECTION 7.7 L’Hôpital’s Rule 383
THEOREM 1 L’Hôpital’s Rule Assume that f (x) and g(x) are differentiable on an
open interval containing a and that
f (a) = g(a) = 0
Also assume that g ′ (x) ̸= 0 (except possibly at a). Then
f (x)
lim
x→a g(x) = lim f ′ (x)
x→a g ′ (x)
if the limit on the right exists or is infinite (∞ or −∞). This conclusion also holds if
f (x) and g(x) are differentiable for x near (but not equal to) a and
lim f (x) =±∞ and lim g(x) =±∞
x→a x→a
Furthermore, this rule if valid for one-sided limits.
EXAMPLE 1 Use L’Hôpital’s Rule to evaluate lim
x→2
x 3 − 8
x 4 + 2x − 20 .
Solution Let f (x) = x 3 − 8 and g(x) = x 4 + 2x − 20. Both f and g are differentiable
and f (x)/g(x) is indeterminate of type 0/0 at a = 2 because f(2) = g(2) = 0:
• Numerator: f(2) = 2 3 − 1 = 0
• Denominator: g(1) = 2 4 + 2(2) − 20 = 0
CAUTION When using L’Hˆopital’s Rule, be
sure to take the derivative of the numerator
and denominator separately:
f (x)
lim
x→a g(x) = lim f ′ (x)
x→a g ′ (x)
Do not differentiate the quotient function
f (x)/g(x).
Furthermore, g ′ (x) = 4x 3 + 2 is nonzero near x = 2, so L’Hôpital’s Rule applies. We
may replace the numerator and denominator by their derivatives to obtain
lim
x→2
x 3 − 8
x 4 + 2x − 2 = lim
x→2
EXAMPLE 2 Evaluate lim
x→2
4 − x 2
sin πx .
3x 2
4x 3 + 2 = 3(22 )
4(2 3 ) + 2 = 12
34 = 6
17
Solution The quotient is indeterminate of type 0/0 at x = 2:
• Numerator: 4 − x 2 = 4 − 2 2 = 0
• Denominator: sin πx = sin 2π = 0
The other hypotheses (that f and g are differentiable and g ′ (x) ̸= 0 for x near a = 2) are
also satisfied, so we may apply L’Hôpital’s Rule:
4 − x 2
lim
x→2 sin πx = lim (4 − x 2 ) ′
x→2 (sin πx) ′ = lim
x→2
cos 2 x
EXAMPLE 3 Evaluate lim
x→π/2 1 − sin x .
−2x
π cos πx =
Solution Again, the quotient is indeterminate of type 0/0 at x = π 2 :
(
cos 2 π
)
= 0, 1 − sin π 2
2 = 1 − 1 = 0
−2(2)
π cos 2π = −4
π
The other hypotheses are satisfied, so we may apply L’Hôpital’s Rule:
lim
x→π/2
cos 2 x
1 − sin x = lim
x→π/2
(cos 2 x) ′
(1 − sin x) ′ = lim
x→π/2
−2 cos x sin x
− cos x
} {{ }
L’Hôpital’s Rule
= lim (2 sin x) = 2
x→π/2
} {{ }
Simplify