# Indeterminate forms of Limits

 Examples with detailed solutions and exrcises that solves limits questions related to inderminate forms such as ∞ / ∞, 0 0, ∞ 0, 1 ∞, ∞ o and ∞ - ∞. A second version of L'hopital's rule allows us to replace the limit problem ∞ / ∞ with another simpler problem to solve. Theorem: If lim f(x) = ∞ and lim g(x) = ∞ 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→∞ ln x / x Solution to Example 1: Since limx→∞ ln x = ∞ and limx→∞ x = ∞ The above L'hopital's rule can be used to evalute the given limit question limx→∞ ln x / x = limx→∞ [ d ( ln x ) / dx ] / [ d ( x ) / dx ] = limx→∞ [ 1 / x ] / 1 = 0 Example 2: Find limx→∞ x e -x Solution to Example 2: Note that limx→∞ x = + ∞ and limx→∞ e - x = 0 This is the indeterminate form ∞ . 0. The idea is to convert it into to the indeterminate form ∞ / ∞ limx→∞ x e -x = limx→∞ x / e x Now apply the above L'hopital's theorem limx→+∞ x / e x = limx→+∞ 1 / e x = 0 Example 3: Find limx→∞ ( 1 + 1/x) x Solution to Example 3: The above limit is of the indeterminate form 1 ∞. If we let t = 1 / x the above limit may written as limx→+∞ ( 1 + 1/x) x = limt→0 ( 1 + t) 1 / t Let y = ( 1 + t) 1 / t and find the limit of ln y as t approaches 0 ln y = (1 / t) ln (1 + t) The advantage of using ln y is that now the limit has the indeterminate form 0 / 0 and the first L'hopital's rule can be applied limt→0 (1 / t) ln (1 + t) = limt→ 0 [ d ( ln(1 + t) ) / dt ] / [ d ( t ) / dt ] = limt→0 [ 1 / (1 + t) / 1 = 1 Since the limit of ln y = 1 the limit of y is e 1 = e, hence limx→∞ ( 1 + 1 / x) x = e Example 4: Find the limit limx→0+ (1 / x - 1 / sin x) Solution to Example 4: Note that limx→0+ 1 / x = +∞ and limx→0+ 1 / sin x = +∞ This limit has the indeterminate form ∞ - ∞ and has to be converted to another form by combining 1 / x - 1 / sin x limx→0+ (1 / x - 1 / sin x) = limx→0+ [ (sin x - x) / (x sin x) ] We now have the indeterminate form 0 / 0 and we can use the L'hopital's theorem. limx→0+ [ (sin x - x) / (x sin x) ] = limx→0+ [ (cos x - 1) / (sin x + x cos x) ] We have the indeterminate form 0 / 0 and use the L'hopital's theorem again. limx→0+ [ (sin x - x) / (x sin x) ] = limx→0+ [ (cos x - 1) / (sin x + x cos x) ] = limx→0+ [ (-sin x) / (cos x + cos x - x sin x) ] = 0 / 2 = 0 Example 5: Find the limit limx→ 0+ x x Solution to Example 5: We have the indeterminate form 00. Let y = x x and ln y = ln (x x) = x ln x. Let us now find the limit of ln y limx→0+ ln y = limx→0+ x ln x The above limit has the indeterminate form 0 . ∞. We have convert it as follows limx→0+ x ln x = limx→0+ ln x / (1 / x) It now has the indeterminate form ∞ / ∞ and we can use the L'hopital's theorem limx→0+ ln x / (1 / x) = limx→0+ (1 / x) / (- 1 / x 2) = limx→0+ -x = 0 The limit of ln y = 0 and the limit of y = xx is equal to e 0 = 1 Exercises: Find the limits 1. limx→∞ (ln x) 1/x 2. limx→∞ [ ln x - ln (1 + x) ] 3. limx→∞ x / e x 4. limx→0+ x sin xSolutions to Above Exercises: Find the limits 1. 1 2. 0 3. 0 4. 1 More on limits Calculus Tutorials and Problems