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 lim_{x→a}, lim_{x→a+}, lim_{x→a-}, lim_{x→+ ∞} or lim_{x→ - ∞}.

__Example 1:__ Find the limit lim_{x→0} sin x / x

__Solution to Example 1:__

Since

lim_{x→0} sin x = 0

and

lim_{x→0} x = 0

L'hopital's rule can be used to evalute the above limit as follows

lim_{x→0} sin x / x = lim_{x→0} [ d ( sin x ) / dx ] / [ d ( x ) / dx ]

= lim_{x→0} cos x / 1 = 1

__Example 2:__ Find the limit lim_{x→0} ( e^{ x} - 1 ) / x

__Solution to Example 2:__

Note that

lim_{x→0} ( e^{ x} - 1 ) = 0

and

lim_{x→0} x = 0

We can use L'hopital's rule to calculate the given limit as follows

lim_{x→0} ( e^{ x} - 1 ) / x = lim_{x→0} [ d ( e^{ x} - 1 ) / dx ] / [ d ( x ) / dx ]

= lim_{x→0} e^{ x} / 1 = 1

__Example 3:__ Find the limit lim_{x→1} ( x^{ 2} - 1 ) / (x - 1)

__Solution to Example 3:__

Since the limit of the numerator

lim_{x→1} ( x^{ 2} - 1 ) = 0

and that of the denominator

lim_{x→1} x - 1 = 0

are both equal to zero, we can use L'hopital's rule to calculate limit

lim_{x→1} ( x^{ 2} - 1 ) / (x - 1) = lim_{x→1} [ d ( x^{ 2} - 1 ) / dx ] / [ d ( x - 1 ) / dx ]

= lim_{x→1} 2 x / 1 = 2

Note that the same limit may be calculated by first factoring as follows

lim_{x→1} ( x^{ 2} - 1 ) / (x - 1)

= lim_{x→1} ( x - 1 )(x + 1) / (x - 1) =

= lim_{x→1} (x + 1) / 1 = 2

__Example 4:__ Find the limit lim_{x→2} ln(x - 1 ) / (x - 2)

__Solution to Example 4:__

Limit of numerator

lim_{x→2} ln(x - 1) = 0

Limit of denominator

lim_{x→2} x - 1 = 0

Both limits are equal to zero, L'hopital's rule may be used

lim_{x→2} ln(x - 1) / (x - 2) = lim_{x→2} [ d ( ln(x - 1) ) / dx ] / [ d ( x - 2 ) / dx ]

= lim_{x→2} [ 1 / (x-1) ] / 1 = 1

__Example 5:__ Find the limit lim_{x→0} (1 - cos x ) / 6 x^{ 2}

__Solution to Example 5:__

Limit of numerator and denominator

lim_{x→0} 1 - cos x = 0

lim_{x→0} 6 x^{ 2} = 0

L'hopital's rule may be used

lim_{x→0} (1 - cos x ) / 6 x^{ 2}
= lim_{x→0} [ sin x ] / [ 12 x ]

The new limit is also indeterminate 0/0 and we may apply l'hopital's theorem a second time

lim_{x→0} (1 - cos x ) / 6 x^{ 2}
= lim_{x→0} [ sin x ] / [ 12 x ]

= lim_{x→0} cos x / 12 = 1 / 12

__Exercises:__ Find the limits

1. lim_{x→0} (sin 4x / sin 2x)

2. lim_{x→0} tan x / x

3. lim_{x→1} ln x / (3x - 3)

4. lim_{x→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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