# 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 - ∞, thenwhere lim stands for lim

_{x→a}, lim

_{x→a+}, lim

_{x→a-}, lim

_{x→+ ∞}or lim

_{x→ - ∞}.

## Examples with Detailed Solutions
## Example 1Find 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 2Find 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 3Find 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 4Find 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 5Find 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
## ExercisesFind the limits1. 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 Exercises1. 22. 1 3. 1 / 3 4. 1 / 2
## More References and linksCalculus Tutorials and ProblemsLimits of Absolute Value Functions Questions |