L'hopital's Rule And The Indeterminate forms 0 / 0



L'hopital's rule allows us to replace a limit problem with another that may be simpler to solve.

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) ]

lim stands for limx→a, limx→a+, limx→a-, limx→+ ∞ or limx→ - ∞.

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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Updated: 2 April 2013

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