L'Hopital's Rule and The Indeterminate Forms 0 / 0 of Limits in Calculus

L'Hopital's theorem allows us to replace a limit problem with another that may be simpler to solve. Several examples are presented along with their solutions and detailed explanations.
L'Hopital's Theorem: If lim f(x) = 0 and lim g(x) = 0 and if lim [ f'(x) / g'(x) ] has a finite value L , or is of the form + ∞ or - ∞, then

lim [ f(x) / g(x) ] = lim [ f'(x) / g'(x) ]

where lim stands for lim
x→a , lim x→a+ , lim x→a- , lim x→+ ∞ or lim x→ - ∞ .

Example 1: Find the limit limx→0 sin x / x

Solution to Example 1:
Since
limx→0 sin x = 0
and
limx→0 x = 0
L'Hopital's rule can be used to evalute the above limit as follows
limx→0 sin x / x = limx→0 [ d ( sin x ) / dx ] / [ d ( x ) / dx ]
= limx→0 cos x / 1 = 1


Example 2: Find the limit limx→0 ( e x - 1 ) / x

Solution to Example 2:
Note that
limx→0 ( e x - 1 ) = 0
and
limx→0 x = 0
We can use L'Hopital's rule to calculate the given limit as follows
limx→0 ( e x - 1 ) / x = limx→0 [ d ( e x - 1 ) / dx ] / [ d ( x ) / dx ]
= limx→0 e x / 1 = 1


Example 3: Find the limit limx→1 ( x 2 - 1 ) / (x - 1)

Solution to Example 3:
Since the limit of the numerator
limx→1 ( x 2 - 1 ) = 0
and that of the denominator
limx→1 x - 1 = 0
are both equal to zero, we can use L'Hopital's rule to calculate limit
limx→1 ( x 2 - 1 ) / (x - 1) = limx→1 [ d ( x 2 - 1 ) / dx ] / [ d ( x - 1 ) / dx ]
= limx→1 2 x / 1 = 2
Note that the same limit may be calculated by first factoring as follows
limx→1 ( x 2 - 1 ) / (x - 1)
= limx→1 ( x - 1 )(x + 1) / (x - 1) =
= limx→1 (x + 1) / 1 = 2


Example 4: Find the limit limx→2 ln(x - 1 ) / (x - 2)

Solution to Example 4:
Limit of numerator
limx→2 ln(x - 1) = 0
Limit of denominator
limx→2 x - 1 = 0
Both limits are equal to zero, L'Hopital's rule may be used
limx→2 ln(x - 1) / (x - 2) = limx→2 [ d ( ln(x - 1) ) / dx ] / [ d ( x - 2 ) / dx ]
= limx→2 [ 1 / (x-1) ] / 1 = 1


Example 5: Find the limit limx→0 (1 - cos x ) / 6 x 2

Solution to Example 5:
Limit of numerator and denominator
limx→0 1 - cos x = 0
limx→0 6 x 2 = 0
L'Hopital's rule may be used
limx→0 (1 - cos x ) / 6 x 2 = limx→0 [ sin x ] / [ 12 x ]
The new limit is also indeterminate 0/0 and we may apply L'Hopital's theorem a second time
limx→0 (1 - cos x ) / 6 x 2 = limx→0 [ sin x ] / [ 12 x ]
= limx→0 cos x / 12 = 1 / 12


Exercises: Find the limits
1. limx→0 (sin 4x / sin 2x)
2. limx→0 tan x / x
3. limx→1 ln x / (3x - 3)
4. limx→0 ( e x - 1 ) / sin 2 x

Solutions to Above Exercises:
1. 2
2. 1
3. 1 / 3
4. 1 / 2

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