L Hôpital s Rule In this note we will evaluate the its of some indeterminate forms using L Hôpital s Rule. Indeterminate Forms and 0 0 f() Suppose a f() = 0 and a g() = 0. Then a g() the indeterminate form of type 0 0. f() Suppose f() = and g() =. Then a a a g() the indeterminate form of type. Note that a can represent a finite real number or + or. may or may not eist and it is called may or may not eist and it is called L Hôpital s Rule: Suppose f and g are differentiable on an open interval containing a and g f() () 0 on that interval ecept possibly for a. Also suppose that a g() is an indeterminate form 0 0 or. Then f() a g() = f () a g () provided that the it on the right side eists or is ±. 5 3 3 + 6 Eample. Evaluate 4 3 + 0. Solution. Since 5 3 3 + 6 = 0 and 4 3 + 0 = 0 we can apply L Hôpital s Rule. So Eample. Evaluate 0 + 5 3 7 + 4. 5 3 3 + 6 4 3 + 0 = 5 6 + 6 8 3 = 4 3 Solution. Since 0 + 5 = and 3 7 + 4 = we can apply L Hôpital s Rule. So 0 + 5 3 7 + 4 = 0 6 7 = 0
e Eample 3. Evaluate cos. Solution. We can NOT apply L Hôpital s Rule because e = and cos = 0. Therefore e cos = Eample 4. Find cos e. Solution. Since cos = 0 and e = 0 we can apply L Hôpital s Rule. So cos e = sin = 0 e 7 Eample 5. Evaluate + sin( ). Solution. Since 7 = 0 and sin( ) = 0 we can apply L Hôpital s Rule. So + + 7 7 + sin( ) = + cos( ) = 7 + cos( ) = Eample 6. Find 3 ln(5 + 3) ln( + 4). Solution. Since 3 ln(5+3) = and ln(+4) = we can apply L Hôpital s Rule. So 3 ln(5 + 3) ln( + 4) = 5 (5+3) (+4) 5 + 60 = 0 + 6 = 5 0 = 3 e Eample 7. Find + sin. Solution. Since e = 0 and sin = 0 we can apply L Hôpital s Rule. So + + e + sin = + e sin + cos As you see e = 0 and sin + cos = 0, so we need to reapply L Hôpital s Rule: + + e e = + sin + sin + cos = e + cos + cos sin =
Indeterminate Form 0 If a f() = 0 and a g() = (or ) then it is not clear what the value of a f()g(), if any, will be. This is called the indeterminate form of type 0. We can convert this type into an indeterminate form of 0 0 or a quotient by writing the product fg as fg = f g or fg = g f Eample 8. Evaluate ln. + Solution. Since = 0 and ln =, the it is an indeterminate form of type + + 0. First we convert this product into the following quotient ln ln = + + where the right side is an indeterminate form of. Then using L Hôpital s Rule we have: ln ln = + + = + = + = 0 Eample 9. Evaluate tan(/). Solution. Since = and tan(/) = 0, the it is an indeterminate form of type 0. First we convert this product into the following quotient tan(/) tan(/) = / where the right side is an indeterminate form of 0. Then L Hôpital s Rule implies that: 0 tan(/) tan(/) = / = / sec (/) / = sec (/) = Eample 0. Evaluate 3 e. Solution. It is not difficult to see that the it is an indeterminate form of type 0. We 3 can easily convert it into the quotient that gives an indeterminate form and then e apply L Hôpital s Rule twice we will have: 3 3 e = e 3 = e = 3 e = 3 4e = 0
Indeterminate Form If a f() = and a g() = then the it a [f() g()] is called the indeterminate form of type. We can convert this type into an indeterminate form of 0 0 or by using a common denominator, or factoring out a common factor or rationalization. Eample. Evaluate Solution. Since (π/) (sec tan ). sec = and tan =, the given it is an indeterminate (π/) (π/) form. Here we use a common denominator to convert it into 0 and then we apply 0 L Hôpital s Rule: (π/) (sec tan ) = (π/) ( cos sin cos ) = sin (π/) ( cos ) = cos (π/) ( sin ) = 0 Eample. Evaluate ( + 3). Solution. It s easily seen that the given it is of the indeterminate form. We can convert it into an indeterminate form using rationalization and then we will apply L Hôpital s Rule: ( + 3). ( + + 3) ( + + 3) = + 3 Note that to evaluate the it + 3 L Hôpital s Rule or an elementary method. 3 ( + + 3) = 3 ( + +3 ) = 3 +3 = in the denominator we can also use
Eercises. Evaluate the following its. Use L Hôpital s Rule where appropriate.. 3 3 9 3 4. 6 5 4 + 7 + 9 sin 7. + cos. 4 3 + 3 5 + 8 e 5. 8. sin sin 3. 3 4 + 3 3 3 5 + + 6. + sin 3 9. e 5 e 3 0. ln.. + cos 3. 4 + 5 ln 4. sin( ) 5. ln( + ) e ln( 0) 6. 7. ( ) 3 ln(4 + ) 8. + ln( + ) e 4 9. e 3 + 0. e e 5. e ( ). sin ln + 5. (csc cot ) 3. csc 4. sin( 4 ) 6. ( + ) 7. +( e ) Final Answers.. 9 4. 3. 5 4 7. 8. 9. 0 4. 0 5. 6. 3 0... 3. 0 4. 5. 0 6. 7. 8. 9. 0. 5. 0. 0 3. 4. 4 5. 0 6. 0 7.