L Hopital s Rule We will use our knowledge of derivatives in order to evaluate its that produce indeterminate forms.
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 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 c yields an 0 0 or, then.. **This will also work with indeterminate forms,,
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
Section 1.3 P. 68 (50 58 even) (67-70) DO ALL PROBLEMS WITH L HOPITAL!! Check all answers with calculator!!
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 3 7x 2 + 1 x 7x 2 + 1 3x x 7x 2 + 1 3
One Sided Limits When evaluating one sided its your first step is always direct SUBSTITUTION. If you get an indeterminate form, use L Hopital. Sometimes, after using L Hopital one or more times, you will end up with #. This means 0 you have a VERTICAL ASYMPTOTE! Then, evaluate the 0 from the left or right to determine if it approaches + or.
What do we ALWAYS have when a it is # 0? Vertical Asymptote! 5 = x 0 + x 5 = x 0 x 5 x 0 + x 5 x 0 x = =
Example Using One Sided Limits. sin x x 0 + x 2 First substitute 0 in as normal and you get an indeterminate form Applying L Hopitals you get. cos x x 0 + 2x 0 0. Now, when plugging in, you get 1 0. (Vertical asymptote). You must evaluate 0 from the right Limit = +
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 0. You must evaluate 0 from the left 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
Evaluating other Indeterminate Forms If, when evaluating a it you do not get an indeterminate form, then you cannot apply L Hopital s. x 3 x 2 x 3 BUT, if you can manipulate the equation, so that you do get an Indeterminate Form, you CAN use L Hopital s. x 3 x 2 (x 3) x 3(x 3) Note: You would use L hopital s twice and the second time you would get a vertical asymptote. It would make no sense to use L Hopital s in this situation. (But you could).
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: 0 We want to rewrite function so it will equal 0 0 or. (Then we can use L Hopital s). 1. If we rewrite the equation so it results in. We now have 0 0 x c f(x) = 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
When we plug in, we get 0 Now, the it = so you can apply L Hopitals...
Graph of y = ln x ln 1 = x 0 ln x = ln x = x
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 Thus
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 lnx = lnx x 0 x 0 + *It is okay to write lnx because by definition, you only evaluate the it within the x 0 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