L Hopital s Rule. We will use our knowledge of derivatives in order to evaluate limits that produce indeterminate forms.

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L Hopital s Rule We will use our knowledge of derivatives in order to evaluate its that produce indeterminate forms.

Indeterminate Limits

Main Idea x c f x g x If, when taking the it as x c, you get an INDETERMINATE FORM.. 0 0 or Then, take the derivative of the top and bottom SEPARATELY and reevaluate the it at c.

Example: x 0 sin x x x 0 cos x 1 Thus the it is 1. **Note that you do not use Quotient Rule!!**

Example: x 1 x 1 x 1 x 1 1 2 x 1 2 1 Thus the it is 1. (Check with calculator) 2 **Note that you do not use Quotient Rule!!**

If the it of f x g(x) INDETERMINATE form as x approaches a yields an 0 0 or, then.. x a f x g x = x a f x g x **This will also work with indeterminate forms,,

Proof of the Special Case

Example 1... Now verify on your calculator.

Solve the following Limits with L Hopitals. x 3 x 1 x 2 +2x 3 x 1 x 1 x 2 +2x 3

Since direct substitution yields., you can apply L Hopital s

Since this results in. you can apply L Hopital s This yields, so you can apply L Hopital s a second time. 0 x x 2 e x = 0

Do the following Limits x 3x 7x 2 + 1 x 3x 2 7x 2 + 1 x 7x 2 + 1 3x x 7x 2 + 1 3

Homework L Hopital s Rule Excercises: (1-8)(13-16)

Example Using One Sided Limits. sin x x 0 + x 2 First substitute 0 in as normal and you get an indeterminate form 0 0. Applying L Hopitals you get. Now, when plugging in, you get 1. (Vertical asymptote). 0 Limit = + cos x x 0 + 2x You must evaluate 0 from the right

sin x x 0 x 2 First substitute 0 in as normal and you get an indeterminate form 0 0. Applying L Hopitals you get. x 0 cos x 2x Now, when plugging in, you get 1. You must evaluate 0 from the left 0 Limit = -

End of Day 1 L Hopitals Hw: Finney Orange Book L Hopital Worksheet (1-9) (11, 12) P. 88 (9-12) Tell me it as x ONLY. P. 205 (20-24)(32-34) Tomorrow: discuss more complicated forms of L Hopital s and putting them into an Indeterminate Form..

Other Indeterminate Forms How do we find a it if, when evaluating, we get something like: f(x) = 0 x c

Special Indeterminate Forms 0, 1, 0 0, 0, If you get one of the following forms when evaluating the it, you can generally manipulate in order to put into the forms 0 0 or.

Indeterminate Form: f(x) = 0 x c 0 We want to rewrite function so it will equal 0 or. 0 (Then we can use L Hopital s). 1. If we rewrite the equation so it results in. We now have 0 0 and we can use L Hopitals. 0 1 2. Or, rewrite the equation so it results in. Which is equivalent to. So we can apply L Hopitals. 1 0

Indeterminate Form 0 When we plug in, we get 0 Now, the it = so you can apply L Hopitals...

x c f x = When dealing with this indeterminate form, you will set the it equal to y and take the Natural Log. y = 1 1 y = x c f x ln y = ln 1 This can be rewritten as ln 1. This equals 0 Like the previous example now you can rewrite to put in indeterminate forms: 0 0 or.

To solve, set the it equal to y.

Since ln is continuous, we know ln() = (ln) Example: ln x 2 x 3 = x 3 ln x 2 ln y = x ln 1 + 1 x x

Can we rewrite to put in 0 0 form? Now you can apply L Hopital s..

Now Use L Hopital s = x 1 1 + 1/x 1/x 1/x This is not our answer..

1 Set equal to Y and take ln 0 0 0 Use L Hopitals to determine and set equal to LN Y. Solve for Y and since Y = Limit, you have your answer.

Homework : L Hopital and Limit Worksheet (17, 18, 21, 22, 45, 47, 49, 50)

For this form you will also take the natural log. (See slide 22 for more detailed explanation of ln approach). ln 0 0 0 ln0 0 *Note that ln(0) is undefined, but, since we are dealing with its, the x 0 lnx = *Also note that for our class x 0 lnx = x 0 + lnx *It is okay to write x 0 lnx because, for our purposes you only evaluate the it within the domain of the function so you would just default to x 0 +

Now you can use L Hopital s because you have

Now you can rearrange this equation

Indeterminate Form 0 For this form you will also take the natural log to convert to form 0 x x1/x y = x x1/x ln y = ln ( x x 1/x ) = ( x ln x 1/x ) = x 1 x ln x = 0 x ln x x =

ln x = x x ln y = 0 = 1/x = x 1 1 = x x = 0 y = e 0 = 1 x x1/x = 1

1, 0 0, 0 Set equal to Y and take ln 0 0 0 Use L Hopitals to determine and set equal to LN Y. Solve for Y and since Y = Limit, you have your answer.

Homework L Hopital WORKSHEET (23-26, 43, 44) Think about the following indeterminate forms: 1, 0 0, 0 Why are these 3 indeterminate while 1 = and 0 = 0 *Hint think about our step where we take ln *

Which of the following are Indeterminate and Which are not? 1, 0 0, 0, 1 and 0 The easiest way to remember which are indeterminate and which are not is by taking the natural log of all and rewriting.

0 vs. 0

This is the last Indeterminate Form that can be written as 0 0 or Unfortunately this, form does not have a defined method to solve.

Need to rewrite to put in 0 0 or so that we can apply L Hopitals First thing to try, get a common denominator = 0 0 So, you can apply L Hopital s

L Hopital s Gives Us: = 0 0 Simplify to make it easier to apply L Hopitals. By L Hopitals you have.

Natural Log Product/Quotient Rules Which of the following are correct? (Hint, use e and 1 to help you) ln x + y = ln x ln y ln x y = ln x ln y ln x y = ln x + ln y ln x y = ln x ln y

What is the Limit? f x = x c 0 5 = 0 f x = x c 0 0 =? f x = 5 x c 0 = Vertical Asymptote f x = x c =? f x = x c 6 2 = 3 f x = x c 3 = f x = x c 0 0 =? f x = x c 0 = 0 f x = x c + = f x = x c =

Indeterminate 0 0 0 0 0 0 1 Determinate + = 0 = 0 = - = 1 = 1 0 = ±

End of L Hopitals Homework: Orange Finney P. 450 (24,25,26,33,34,42,46)

Relative Rates of Growth A useful extension of L Hopitals is using it to determine which of two functions grows the fastest. If you have two functions, A and B, you can determine which of the two grows the fastest by setting up a fraction and taking the its as x approaches infinity.

If you have two functions, A and B, you can determine which of the two grows the fastest by setting up a fraction and taking the its as x approaches infinity. If. a x b = Then a grows faster than b. a If. = finite # x b Then they grow same rate a If. = 0 x b b grows faster than a.

Which grows faster. e x or x 2? To solve this, take the it of the ratios of the two functions. x e x x 2 You will need to use L Hopital s twice as the first two times you plug in you get. x e x 2 = Thus you find that e x grows faster than x 2.

Which grows faster. x 3 or 2 x? x x x 3 2 x 3x 2 ln2 2 x x 6 x ln2 3 2 x 6x ln2 2 2 x = 0. 2 x grows faster.

So which will eventually grow faster. x 1,000,000 or 2 x