Dr. Z s Math151 Handout #4.7 [L Hôspital s Rule]

Transcription

1 By Doron Zeilberger Dr Z s Math151 Handout #47 [L Hôspital s Rule] Problem Type 471 : Given certain its of certain functions f(x) g(x) at a designated point x = a determine whether the its (at that very same point x = a) of the quotient f(x)/g(x) product f(x)g(x) difference f(x) g(x) and exponentiation f(x) g(x) are indeterminate its Problem 471: Given that f(x) = 0 g(x) = 0 p(x) = q(x) = which of the following are indeterminate forms (a) [f(x)/g(x)] (b) [f(x)p(x)] (c) [p(x) q(x)] (d) [f(x) g(x) ] a For a quotient the it is indeterminate whenever plugging in yields 0/0 or / a f(x) g(x) = 0 0 hence this is an indeterminate form b For a product the it is indeterminate whenever plugging in yields 0 or 0 (and of course 0 or 0) b [f(x)p(x)] = f(x) p(x) = 0 hence the it is indeterminate c The it of a difference is indeterminate whenever it is of the type (or ( ) ( )) c [p(x) q(x)] = hence the it is indeterminate

2 d The it of an exponentiation is indeterminate whenever it is of the type or 1 d [f(x)g(x) ] = 0 0 hence the it is indeterminate

3 Problem Type 472 : Use L Hospital s rule if appropriate to evaluate T OP (x) BOT (x) Problem 472: Use L Hospital s rule if appropriate to evaluate x 8 1 x 1 x First plug-in x = a into T OP (x)/bot (x) and see whether T OP (a)/bot (a) yields 0/0 or / If it does then L Hospital s rule is applicable 1 Plugging-in x = 1 into x8 1 x 3 1 gives 0/0 so L Hospital s rule is applicable 2 Invoke L Hospital s rule 2 T OP (x) BOT (x) = T OP (x) BOT (x) x 8 1 x 1 x 3 1 = (x 8 1) x 1 (x 3 1) = 8x 7 x 1 3x 2 If you still get an indeterminate form (in this example you don t) keep doing it until you get a doable it 3 Evaluate the it by simplifying and plugging-in 3 Ans: 8/3 8x 5 = x 1 3 = = 8 3

4 Problem Type 473 : Same as 472 but now you have to use L Hospital s rule more than once Problem 473: Use L Hospital s rule if appropriate to evaluate 1 cos x x 0 x 2 1 First plug-in x = a into T OP (x)/bot (x) and see whether T OP (a)/bot (a) yields 0/0 or / If it does then L Hospital s rule is applicable 1 Plugging-in x = 0 into 1 cos x x gives 2 0/0 so L Hospital s rule is applicable 2 Invoke L Hospital s rule 2 T OP (x) BOT (x) = T OP (x) BOT (x) 1 cos x (1 cos x) sin x x 0 x 2 = x 0 (x 2 ) = x 0 2x If you still get an indeterminate form (in this example you do!) keep doing it until you get a doable it Now plugging-in x = 0 still yields 0/0 so we have to do L Hospital again (sin x) = x 0 (2x) = x 0 cos x 2 3 Evaluate the it by simplifying (if necessary) and plugging-in 3 = cos 0 2 = 1 2 Ans: 1/2

5 Problem Type 474 : Use L Hosptial s rule (or any other method) to evaluate [ Expression 1(x) Expression 2 (x) ] where one of the expressions is a radical (ie involves the square-root sign) and plugging-in gives Problem 474: Use L Hospital s rule or any other method to evaluate [ x 2 + 3x x] 1 Multiply top and bottom by the conjugate Expression 1 (x)+expression 2 (x) and simplify as much as you can using (a b)(a + b) = a 2 b 2 1 [ x 2 + 3x x] = ( x 2 + 3x x)( x 2 + 3x + x) = x2 + 3x + x ( x 2 + 3x) 2 x 2 x2 + 3x + x (x 2 + 3x) x 2 x2 + 3x + x = 3x x2 + 3x + x = 2 If you can get by without L Hospital s rule don t bother using it (it may be complicated) Try to use any other rules 2 In this case you can use the only the leading term counts as x what I call forget about the little ones = 3x ( x 2 + x) where we ignored 3x in view of the much more important x 2 and we get 3x = x + x = 3 2 = 3 2 Ans: 3/2

6 Problem Type 475 : Use L Hosptial s rule (or any other method) to evaluate Expression 1(x) 1/Expression 2(x) where plugging in will give 0 Problem 475: Use L Hospital s rule or any other method to evaluate x1/2x 1 Taking natural logarithms evaluate instead ln(expression 1 (x)) ln(expression 2 (x)) using L Hospital s rule if necessary 1 ( ln x 1/2x) ln x = 2x = (ln x) (2x) 1/x = 2 = 1 2x = 0 2 But what you got now is not the answer but the log-natural of the answer To get the answer to the problem you have to undo the effect of ln by exponentiating So the final answer is exp(above Limit) 2 Ans: e 0 = 1 A Problem from a Previous Final (Spring 2008 #5 (10 points)) Evaluate the given its a) (4 points) b) (6 points) x 3 + 2x x + e x x x 0 x2 ln x +

7 Solutions a) This is an indeterminate form / We have to use L Hôpital s rule three times! x 3 + 2x x + e x x = 3x x + e x 1 = Now the top is not and plugging it in we get = 6 e = 6 = 0 Officially we have to say because e x = Ans to a): 0 6x x + e x = x + b) First we need to simplify to make life easier using ln x = ln x 1/2 = (1/2) ln x 6 e x x 2 ln x = (1/2)x 2 ln x x 0 + x 0 + Next we need to bring into a form A/B by rewriting x 2 as 1/x 2 x 0 + ln x 2x 2 Now we are ready to apply L Hôpital s rule x 0 + Using the algebra of exponents we get x 0 + x 1 x 1 3 = 4x 3 x ln x 1/x = 2x 2 x 0 + 4x 3 x 2 = x = 02 4 = 0 Ans to b): 0