# L'hopital's Rule And The Indeterminate forms 0 / 0

 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 limx→a, limx→a+, limx→a-, limx→+ ∞ or limx→ - ∞. Example 1: Find the limit limx→0 sin x / x Solution 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 2: Find 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 3: Find 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 4: Find 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 5: Find the limit limx→0 (1 - cos x ) / 6 x 2 Solution 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 Exercises: Find the limits 1. 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 Exercises: 1. 2 2. 1 3. 1 / 3 4. 1 / 2 More on limits Calculus Tutorials and Problems