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 limx→0 sin x / xSolution 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 2Find 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 3Find 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 4Find 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 5Find the limit limx→0 (1 - cos x ) / 6 x 2Solution 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 ExercisesFind the limits1. 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 Exercises1. 22. 1 3. 1 / 3 4. 1 / 2
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