Section 4.5: Indeterminate Forms and L Hospital s Rule Practice HW from Stewart Tetbook (not to hand in) p. # 5-9 odd In this section, we want to be able to calculate its that give us indeterminate forms such as and. In Section.5, we learned techniques for evaluating these types of it which we review in the following eamples. 9 Eample : Evaluate Eample : Evaluate 4
However, the techniques of Eamples and do not work well if we evaluate a it such as e For its of this type, L Hopital s rule is useful. L Hopital s Rule Let f and g be differentiable functions where g ( ) near a (ecept possible at a). If f ( ) a g ( ) produces the indeterminate forms,,, or provided the it eists. f ( ) a g( ), then f ( ) a g ( ) Note: L Hopital s rule, along as the required indeterminate form is produced, can be applied as many times as necessary to find the it.
Eample : Use L Hopital s rule to evaluate 9 Eample 4: Use L Hopital s rule to evaluate 4.
4 Eample 5: Evaluate e Note! We cannot apply L Hopital s rule if the it does not produce an indeterminant form,,, or. Eample 6: Evaluate
5 a a Helpful Fact: An epression of the form, where a, is infinite, that is, evaluates to or. Eample 7: Evaluate e For this problem, first realize that if we directly substitute in, we get the following: e () e e () Since we have the indeterminant form, L Hopital s rule applies. Applying the rule once gives e e e It is tempting to use L Hopital s again. However, direct substitution results in the following () e e e () () The result is not an indeterminate form, but a non-zero number divided by, which results in an infinite it. To see what type of infinite behavior occurs, one can eample that as gets closer to from the right, the numerator e while the denominator steadily becomes a smaller and smaller positive number. Taking a finite number and dividing by a small positive e gives a large number. Thus. If you still have trouble convincing yourself of this, using a plot chart such as the following can be helpful. Approaches from the right Obviously, it can be seen that e. e (.). e.8 (.).. 54.5. 54.4 e. Thus, in summary we see that. e e
6 Other Types of Indeterminant Forms Note: For some functions where the it does not initially appear to as an indeterminant,,, or. It may be possible to use algebraic techniques to convert the function one of the indeterminants,,, or before using L Hopital s rule. Indeterminant Products Given the product of two functions f g, an indeterminant of the type or results (this is not necessarily zero!). To solve this problem, either write the product as f g or and evaluate the it. / g / f Eample 8: Evaluate e
7 Eample 9: Evaluate ln As, and ln (if you do not believe this, take your calculator and compute the natural logarithm of some very small positive numbers, like ln(. ), ln(.), ln(.), ln(.), etc. Hence, we have the indeterminant form. The trick of this problem is write the it as ln (Note that ln ln ln ln ) In this form, we have the indeterminant form This gives and can indeed use L Hopital s Rule. ln ln (Note that d d (ln ) and d d d d ( ) ) (Take the reciprocal of the denominator and multiply) (Simplify) () (Evaluate the it) Thus, ln
8 Indeterminate Differences Get an indeterminate of the form (this is not necessarily zero!). Usually, it is best to find a common factor or find a common denominator to convert it into a form where L Hopital s rule can be used. Eample : Evaluate csc cot Indeterminate Powers Result in indeterminate,, or. The natural logarithm is a useful too to write a it of this type in a form that L Hopital s rule can be used.
9 Eample : Evaluate ( e ) On this problem, if we substitute in directly into the it, we obtain ( e ) ( e ) ( ) logarithms to solve this it. We start by setting y, and indeterminate power. We can use ( e ) Taking the natural logarithm of both sides gives or ln y ln[ ( e ) ln y [ln( e ) Using the eponent property of logarithms ln k ln u gives ln y [ ln( e )] or ln( e ) ln y ln( e ) ln( e ) Direct substitution for the right hand side yields which says L Hopital s rule can be used. This gives u k ] ] ln(). ln y e e ln( e ) ( e ) e e e (Use the chain rule for the ln function to differentiate the numerator) (Simplify) (Sub to find the it) (continued on net page)
Hence, we have found that ln y Recall by definition that ln k loge k. Hence, the equation becomes log e y By definition, recall that log a s t means that t s a. Thus, our equation becomes However, recall above that we set y y e ( e ) Hence, substituting for y, we obtain the following answer., which was our original it. ( e ) e