Calculus II: The derivative : applications
l'Hospital's rule

1. L'Hospital's rule

Let f(x) and g(x) two functions continuous on an closed interval [a,b] and differentiable on the open interval ]a,b[. In addition let c a real in the interval [a,b] such as:

lim f(x) = 0
x → c

and

lim g(x) = 0
x → c

So that
lim f(x)/g(x) = 0/0: Undetermined
x → c

We construct two new functions:
F(x) = f(x) for a < x < b with x ≠ c and F(c) = 0,
G(x) = g(x) for a < x < b with x ≠ c and G(c) = 0,

Therefore

1. For x > c:
F(x) and G(x) are continuous on the interval [c,x] and differentiable on ]c,x[.

According to the generalized mean value theorem there exists a real r on ]c,x[ such that:

(F(x) - F(c))/(G(x) - G(c)) = f'(r)/g'(r)
That is
F(x) /G(x) = f'(r)/g'(r)

Since F(x) = f(x) and G(x) = g(x) in the whole [a,b]\{c}, we have :

F(x) /G(x) = f(x)/g(x) = f'(r)/g'(r)

When x tends to c⁺, so does r. Taking the limits

lim f(x)/g(x)
x → c⁺
= lim f'(r)/g'(r)
r → c⁺

Since x is any real in [a,b], we obtain:

lim f(x)/g(x)	= lim f '(x)/g'(x)
x → c+	  x →   c+

2. Similarly,

For x &t; c:
(F(c) - F(x))/(G(c) - G(x)) = - f(x)/(- g(x)) = f(x)/g(x) = f'(r)/g'(r)

We find

lim f(x)/g(x)	= lim f '(x)/g'(x)
x → c-	    x → c-

f(x) and g(x) are continuous on the whole interval [a,b], so

lim f(x)/g(x) =	lim f '(x)/g '(x)
x → c	x → c

This theorem is proved for the form 0/0. By changing x(x) in 1/f(x) and g(x) in 1/g(x), the theorem holds for the limits at ±∞. so

whether we have 0/0 or ∞./∞, we can use l'Hospital's rule.

What l'Hôpital tells us?

2. Examples:

2.1. Example 1:

f(x) = sin(x)
g(x) = x

lim (f(x)/g(x)) = 0/0 Undetermined: L'Hospital's rule holds.
x → 0

So

lim sin(x)/x =	lim cos(x)/1 = 1/1 = 1
x → 0	x → 0

lim sin(x)/x = 1
x → 0

2.2. Example 2:

f(x) = cos(x) - 1
g(x) = x

lim (f(x)/g(x)) = 0/0: Undetermined. L'Hospital's rule holds.
x → 0

So

lim (cos(x) - 1)/x =	lim (- sin(x)/1) = 0
x → 0	x → 0

lim (cos(x) - 1)/x = 0
x → 0