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 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→ - ∞}.

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 ( x² - 1 ) / (x - 1)
Solution to Example 3:
Since the limit of the numerator
lim_x→1 ( x² - 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 ( x² - 1 ) / (x - 1) = lim_x→1 [ d ( x² - 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 ( x² - 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 x²
Solution to Example 5:
Limit of numerator and denominator
lim_x→0 1 - cos x = 0
lim_x→0 6 x² = 0
L'Hopital's rule may be used
lim_x→0 (1 - cos x ) / 6 x² = 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 x² = 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