L’Hospital’s rule
(Stewart, § 4.4)
The rule is named after the French mathematician Guillaume François Antoine, Marquis de l’Hospital
(later also spelled as l’Hôpital), who lived from 1661 till 1704. He published the rule in 1696 in his
textbook Analyse des infiniment petits on differential calculus, however the rule was discovered in 1694
by the Swiss mathematician Johann Bernoulli (1667-1748).
The rule is based on the definition of the derivative of a function (see: Stewart, § 2.7):
f 0 (a) = x→a
lim
f (x) − f (a)
.
x−a
This can be used for instance to find that:
sin x − sin 0
d
sin x
= cos 0 = 1,
= lim
=
sin x
lim
x→0
x→0 x
x−0
dx
x=0
1 1
arctan x
arctan x − arctan 0
d
=
=
lim
= lim
=
arctan x
=1
2
x→0
x→0
x
x−0
dx
1 + x x=0 1 + 0
x=0
and
ln x
1 1
ln x − ln 1
d
lim
= = = 1.
= lim
=
ln x
x→1 x − 1
x→1
x−1
dx
x x=1 1
x=1
This idea can be generalized as follows: suppose that f (a) = g(a) = 0, then we have
f (x) − f (a)
f (x)
= x→a
lim
= x→a
lim
lim
x→a g(x)
g(x) − g(a)
f (x)−f (a)
x−a
g(x)−g(a)
x−a
f (x)−f (a)
x−a
g(x)−g(a)
lim x−a
x→a
lim
=
x→a
=
f 0 (a)
f 0 (x)
=
lim
,
g 0 (a) x→a g 0 (x)
if f 0 and g 0 are both continuous at x = a (see: Stewart, § 2.5).
If lim f (x) = lim g(x) = 0, then we use the substitution x = 1/t to obtain:
x→∞
x→∞
lim
x→∞
f (1/t)
f 0 (1/t) · (−1/t2 )
f 0 (1/t)
f 0 (x)
f (x)
= lim
= lim 0
=
lim
=
lim
.
x→∞ g 0 (x)
t↓0 g(1/t)
t↓0 g (1/t) · (−1/t2 )
t↓0 g 0 (1/t)
g(x)
Some examples:
1
ln x
1/x
= lim
= ,
− 1 x→1 2x
2
1 2
x
x
e − 1 − x − 2x
e −1−x
ex − 1
ex
1
lim
=
lim
=
lim
=
lim
= ,
3
2
x→0
x→0
x→0
x→0 6
x
3x
6x
6
1 2
cos x − 1 + 2 x
− sin x + x
− cos x + 1
sin x
cos x
1
lim
= lim
= lim
= lim
= lim
= ,
4
3
2
x→0
x→0
x→0
x→0 24x
x→0 24
x
4x
12x
24
lim
x→1 x2
and also
1
sin x − x
cos x − 1
− sin x
−0
1
lim
−
= lim
= lim
= lim
=
= 0,
x→0 x sin x
x→0 x cos x + sin x
x→0 2 cos x − x sin x
x→0 x
sin x
2−0
x
1
x ln x − x + 1
ln x + 1 − 1
x ln x
ln x + 1
1
lim
−
= lim
= lim
= lim
= lim
=
x−1
x→1 x − 1
x→1 (x − 1) ln x
x→1 ln x +
x→1 x ln x + x − 1
x→1 ln x + 2
ln x
2
x
and
lim x ln(1+ x1 )
1 x
x ln(1+ x1 )
x→∞
=
lim
e
lim
1
+
=
e
= e1 = e,
x→∞
x→∞
x
since
1
1
1
·
−
ln 1 + x
x2
1
1
1
1+ x1
lim
x
ln
1
+
=
lim
=
lim
=
lim
=
= 1.
1
1
1
x→∞
x→∞
x→∞
x→∞ 1 +
x
1+0
− x2
x
x
L’Hospital’s rule also holds in the cases
h
±∞
±∞
i
instead of
h i
0
0
, but is more difficult to prove. Examples:
x3
3x2
6x
6
=
lim
= lim 2x = lim 2x = 0
2x
2x
x→∞ e
x→∞ 2e
x→∞ 4e
x→∞ 8e
lim x3 e−2x = lim
x→∞
and
2 ln x ·
(ln x)2
= lim
x→∞
x→∞
x
1
lim
1
x
2 ln x
2/x
= lim
= 0.
x→∞
x→∞ 1
x
= lim