Extra Fun: The Indeterminate Forms 1, 0, and 0 0

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1 math 30, day 38 its: l hôpital s rule, part 7 Etra Fun: The Indeterminate Forms, 0, and 0 0 Some of the most interesting its in elementary calculus have the indeterminate forms,0 0, or 0 0. All of these indeterminate it forms arise from functions that have both a variable base and a variable eponent (power). For eample, consider!0 Form: 0 0 Form:!! / Form: 0 We will use logs and l Hôpital s rule to simplify some of these it calculations. General Form The general form of all of these its is [ f ()] g() = y. To!a simplify these its we use the natural log to undo the power. If the eventual it is y (which is unknown to us it s what we are trying to find, then y =!a [ f ()] g(). We take the natural log of both sides here [ f ()] g() is assumed to be positive. y = (!a [ f ()] g() ) As long as f () and g() are continuous, we can switch the order of the log and the it and use log properties y Cont =!a ([ f ()] g() ) y =!a g() ( f ()) At this stage we typically use l Hôpital s rule to find the it, call it L. Then y = L so we must have y = e L. Let s look at some eamples. EXAMPLE Determine (). Notice that this is a 0 0 form. SOLUTION. Let y = (). We want to find y. Using the log process above, y = ( () ) y Cont = () y = y = y l Ho = y = But y = 0 implies y = e 0 =. So () = y =. Wow! EXAMPLE 38.. (Critical Eample). Determine. Notice that this is a form.!

2 math 30, day 38 its: l hôpital s rule, part 8 SOLUTION. Let y =. We want to find y. Using the log process,! apple y =! y Cont =! y =! y =! y l Ho =! y =! y =. But y = implies y = e = e. So = y = e. Double Wow!! In fact, in! some courses you will see that e is defined this way. EXAMPLE (Critical Eample). Determine! /. Notice that this is a 0 form. SOLUTION. Let y =! /. We want to find y. Using the log process, y =! / y Cont =! / y =! y =! y l Ho =! y = 0 = 0. But y = 0 implies y = e 0 =. So! / = y =. Neat! EXAMPLE Determine! (e ) /. Notice that this is a 0 form again. SOLUTION. As usual let y =! (e ) /, so y = (e / Cont ) = (e ) /!! y =! (e ) (e y = )! y l Ho e e =! y = e! e e y l Ho =! e y l Ho =! e e y =.

3 math 30, day 38 its: l hôpital s rule, part 9 But y = implies y = e. So! (e ) / = e. EXAMPLE Determine! [5( )]. Notice that this is a 0 0 form. SOLUTION. Let y = above,! [5( )]. We want to find y. Using the log process y = ( )]! [5( y Cont = [5( )]! y = ) [5( )]! ( [5( )] y =! y l Ho =! 5 ( ) y = 5( )! But y = 0 implies y = e 0 =. So! [5( )] = y =. This is becoming routine. YOU TRY IT Here s a great problem to see if you have mastered these ideas. Determine Determine [sin()]. Problems Answer to you try it 38.3 :. Hint: Use l Hôpital s rule twice.. Some interesting its. Answers not in order: 0, 0, 0,, 4,,,, e, e k,,, and 6, (a)!0 cos 4 cos (d)! e (e) cos (g) e (b)!0 (c)!0 (3) (h) ( )/ (i) k! (j)!0 arctan 4 sin (m) (k)!0 sin 4 3 sec (f ) [(4 9) ( 7)]! (l)!

4 math 30, day 38 its: l hôpital s rule, part 0 Solutions. Make sure to check those stages at which l Hôpital s rule applies. cos 4 cos l Ho 4 sin 4 sin (a)!0 =!0 6 4 = 6. e l Ho e (b)!0 = l Ho =!0 (c) l Ho = =!0 e =. l Ho =!0 6 cos 4 4 cos = = = 0. (d) e = l Ho l Ho!! e =! e =! e = 0 (i.e.,! ) cos (e)! 0 :. l Hôpital s rule does not apply. (f ) (4 9)! = 4. ( 7) =! =! (g) Let y = (3). We want to find y. Using the log process, = ( (3) ) y Cont = (3) y = 3 3 y = y l Ho = y = 3 3 But y = 0 implies y = e 0 =. So (3) = y =. (h) " ": Let y = ( )/, so y = ( )/ y Cont = ( )/ y = ( ) ( ) y = y l Ho = y = y =. But y = implies y = e. So ( )/ = e. (i) " ": Let y = ( k)/, so 4 9 l Ho 4 = 7! =

5 math 30, day 38 its: l hôpital s rule, part y = ( k)/ y Cont = ( k)/ y = ( k) ( k) y = y l Ho = y = k y = k. k k But y = k implies y = e k. So (!0 k)/ = e k. 4 4 arctan 4 l Ho 6 (j) =!0 sin!0 cos = =. sin 4 (k)!0 3 sec = 0 = 0. l Hôpital s rule does not apply. 3 (l)!! 0 (m) " 0 ": Let y =, so :. l Hôpital s rule does not apply. y = y Cont = y = ( ) y = y l Ho = y = 3 y = But y = 0 implies y =. So y = =. 3