TEMPLE MAES 2189 - L’Hˆopital’s Rule and Indeterminate Forms
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L H opital s Rule and Indeterminate Forms Created by Tynan Lazarus October 29 2018 Now that we have the power of the derivative we can use it as a way to compute limits that we didn t have the ability to understand before Early on we could compute limits of rational functions quite easily However we couldn t deal with mixing a ratio of di erent kinds of functions like a polynomial and an exponential L H opital s Rule allows us to evaluate these kinds of limits without much e ort It also allows us to deal with di erent indeterminate forms We will see through some examples just how weird can act and why these indeterminate forms bring about contradictions in our intuition 1 1 The De nition Theorem L H opital s Rule Let f x and g x be di erentiable on an interval I containing a and that g cid 48 a cid 54 0 on I for x cid 54 a Suppose that lim x a f x g x 0 0 or lim x a f x g x Then as long as the limits exist we have that lim x a f x g x lim x a f cid 48 x g cid 48 x There is an analogous version for when a is or What this theorem essentially says is that if you tried to compute the limit of a ratio of functions but you get the indeterminate forms 0 or 0 then you can compute the limit of the ratio of the derivatives of those functions instead However take caution that it is not necessarily a short cut When encountering limits that we have seen before it may be faster to use a di erent technique than L H opital s Rule Also note that we are not taking a quotient rule We just take the derivatives of the top and the bottom of the fraction and leave them there 1 2 How it Works Before we could compute the derivative of sin x or cos x we had to gure out two trig limits We sin x found that lim x 0 know how to compute derivatives we can use L H opital s Rule to check that this is correct 1 by using some geometry tricks with sectors and whatnot Now that we x In order to use L H opital s Rule we need to check that it is in the right form and that we get one of the indeterminate forms required As usual with limits we attempt to just plug in the value and see if we get a number If we did get a real number then we are done Here we can see that if we try to plug in x 0 in the limit we get that lim x 0 which is an indeterminate form Therefore we can apply L H opital s Rule Whenever we do so we will use a L H to denote that we have used the rule and to denote our usual simpli cation So applying L H opital s Rule we sin x x 0 0 get lim x 0 sin x x L H lim x 0 cos x 1 so we can just plug x 0 and get that lim x 0 already know to be true However this second expression is a limit of a continuous function sin x x L H lim x 0 cos x 1 cos 0 1 verifying what we 1 Created by Tynan Lazarus We can do the same with our other trig limit lim x 0 0 that L H opital s Rule even applies 0 which is one of our indeterminate forms L H opital s Rule does apply to this form so we get that If we tried to plug in x 0 we would get 0 cos x 1 x 0 First we have to check cos 0 1 cos x 1 lim x 0 x sin x L H lim x 0 1 sin 0 0 1 3 Examples with Indeterminate Forms 1 3 1 Form 0 0 Question Why is 0 the limit down towards zero and the bottom pulls it up to in nity So who wins 0 indeterminate In general stu 0 and stu 0 0 acts like So the top pulls Let s say we want to compute lim x 2 x 2 x2 4 0 0 Therefore we can apply L H opital s Rule to get We can see that if we try to plug in x 2 we get x 2 x2 4 lim x 2 L H lim x 2 1 2x 1 2 2 1 4 But this is a limit that we could ve computed in the rst week of the course we don t even need the relative canon that is L H opital to swat this little limit Earlier we would ve just factored the bottom and gotten x 2 x2 4 x 2 x 2 x 2 lim x 2 lim x 2 x 2 lim x 2 1 4 1 Either way is just as quick because this is a simple limit For a more interesting example let s try to compute lim x 0 ln sec x 3x2 We see a limit so our rst instinct is to plug in the limiting value x 0 When we do this we get 0 0 This is one of the indeterminate forms that L H opital s Rule can help us with So we use it to get 3 0 2 ln sec 0 ln 1 0 ln sec x lim x 0 3x2 L H lim x 0 sec x sec x tan x 1 6x lim x 0 tan x 6x Trying to take this limit also results in 0 0 So did L H opital fail us Not quite All L H opital tells us is that the limit of the original ratio is that same as the limit of the ratio of the derivatives We got 0 0 which is what L H opital s Rule is designed for so let s use it again Thus we get ln sec x lim x 0 3x2 L H lim x 0 tan x 6x L H lim x 0 sec2 x 6 1 6 Therefore our original limit has a value of 1 6 This problem shows us that you may need to use L H opital s Rule multiple times before we get an answer However we do need to check that we are in the correct indeterminate form each time before we can apply it 2 Created by Tynan Lazarus Form 1 3 2 Question Why is the limit up to in nity and the bottom tries to pull it down to 0 So who wins indeterminate Usually stu acts like and stu goes to 0 So the top pulls Consider the following limit 2x2 e3x Since this ratio is of a polynomial and an exponential function we can t solve it with any of the usual techniques from earlier in the course …

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