# L'Hopital's Rule and The Indeterminate Forms 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

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

f '(x) and g '(x) are the derivatives of f(x) and g(x) respectively.

## Examples with Detailed Solutions

### 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

## More References and links