Some fats you should know that would be onvenient when evaluating a it: When evaluating a it of fration of two funtions, f(x) x a g(x) If f and g are both ontinuous inside an open interval that ontains a (but not neessarily at a), we an atagorize the nature of the it by what the numerator and denominator approahes as x a: Division by a Non-zero number: f(x) f(x) If g(x) as x a, then x a g(x) = x a f(x) x a g(x) x + 4 x 3 x 5 = 7 2 In this example, f(x) = x + 4, g(x) = x 5, sine f and g are both ontinuous funtions, and g(x) as x 3, we an evaluate the it by simply plug in the value: x + 4 x 3 x 5 = 3 + 4 3 5 = 7 2 = 7 2 This result simply means that we an always perform a division by a non-zero real number. Non-zero divided by zero: If f(x) but g(x) as x a, then the it will be either or, or undefined. x 1 x + 3 x 1 In this example, the denominator is g(x) = x 1 while the numerator is f(x) = x + 3. g(x) while f(x) as x 1, therefore the it will either be or. In fat, x 1 x + 3 x 1 =
x + 3 x 1 + x 1 = This is the result of when a non-zero number is divided by a very small number, the result will be a very large number. To put it simply, non-zero divided by zero is infinity. Indeterminate Form: Zero divided by Zero: This is the more interesting ase and usually requires more algebra. What this means is that, as x a, both f(x) and g(x). Understand that a fration will be a small number if the numerator is small (lose to zero), but a fration will be a large number if the denominator is small. In the indeterminate ase of, we have a ompetition between the numerator and the denominator. Sine both the numerator and denominator will approah as x a, We are taking a small number divided by another small number, and the result will depend on whih funtion (numerator or denominator) goes to zero faster. If the numerator wins, meaning that f(x) goes to faster than g(x), the resulting it will be. If the denominator wins, meaning that g(x) goes to faster than f(x), the resulting it will be or. If the two funtions go to at about the same rate, the resulting it ould be any non-zero real number depending on the funtion. Consider the following its: x x = 1 x x 1 + 1 os x = x ln x x 2 2x + 1 = In all three of the above frations, both the numerator and denominator approahes as x approahes a, but these three its have different values. A similar argument an used to approah it problems where f(x) or
g(x) as x a. In this argument the result will still work if we replae any real number a by the or symbol, assuming that f and g are ontinuous in the appropriate region. Infinity divided by Finite: If f(x) ± and g(x) ± as x a or as x, then the it will be or or undefined. ln x 5 = In this example, the numerator is f(x) = ln x, and the denominator is g(x) = 5. f(x) as x, while g(x) 5, the result of this it is. The above means that a number that is extremely large divided by a number that is not as large, the result will still be a large number. Intuitively, infinity divided by finite is still infinite. Finite divided by Infinity: If f(x) ± and g(x) ± as x a or as x, then the it will be x = In this example, the numerator is f(x) =, and the denominator is g(x) = x. f(x) as x, but g(x), so this it is. In other words, finite divided by infinite is textbfindeterminate form: Infinity divided by Infinity: This is another ase where additional algebra will be needed. If both f(x) ± and g(x) ± as x a or x, then the it ould be any real number, or ±, or undefined. ln x x =
x 2 3x + 4 2x 2 + 5 e x x = = 1 2 The situation in this indeterminate ase is similar to the indeterminate ase. A large denominator will make the fration small, while a large numerator will make the fration big, so a ase will depend on whih funtion wins. If the numerator goes to infinity faster than the denominator, the result will be infinity. If the denominator goes to infinity faster than the numerator, the result will be. If the numerator and denominator goes to infinity at about the same rate, the result ould be any non-zero real number depending on the funtion. The following its use some of the fats we just mentioned: Let be any onstant, ln x = if p >, x = p if b > 1 b = x if r < 1, r x = There are other types of indeterminate forms where we have a ompetition between two funtions where the value of one funtion would ause the it to behave one way, but the value of the other funtion would ause the it to behave another way. The result will depend on whih funtion wins, or whih funtion approahes its it value at a faster rate. The following are all indeterminate forms:,,,,, 1, We now introdue a useful theorem that helps us to evaluate the it of an indeterminate form:
L Hospital s Rule: Suppose f and g are differentiable funtions and g (x) for some open interval that ontains a (exept possibly at a). If any one of the following two ondition is satisfied: i) x a f(x) = and x a g(x) = ii) x a f(x) = ± and x a g(x) = ± In other words, we have a it of the indeterminate form or then: f(x) x a g(x) = f (x) x a g (x) The above formula is still valid if a is replaed by or. Before trying to use L Hospital s Rule to evaluate a it, you must first make sure that the it satisfy the riterion for L Hospital s rule. L Hospital s rule an be applied only when we have an indeterminate form or. You may not apply L Hospital s rule if you have something like or Do not try to apply L Hospital s rule for the following it: x 2 We annot apply L Hospital s rule for this it beause, while the denominator, g(x) = x 2, approahes as x, the numerator, f(x) =, does not approah or as x, so this does not fit the hypothesis of L Hospital s rule, hene the rule annot be applied. Another thing to note is that, when applying L Hospital s rule, you will try to evaluate the it by taking the derivative of the numerator and the denominator as individual funtions. Do not try to differentiate the funtion as a single fration.
x x For this it, sine and x as x, this is an indeterminate form. We may apply L Hospital s rule. Sine the derivative of is os x and the derivative of x is 1, we have: x x = os x x 1 = 1 It will be inorret if you try to differential the funtion x when trying to apply L Hospital s rule. x x x os x x x 2 ln x Evaluate: x as a single fration Ans: As x, ln x, and x, so we may apply L Hospital s rule: ln x 1/x = x 1/(2 x) = 2 x x = 2 = x When we apply L Hospital s rule, we are trying to turn an indeterminate form into a form that is no longer indeterminate. One we are able to do that, we will hopefully be able to find out the behavior of the it. Evaluate the it: x + x ln x Ans: This it is one of the indeterminate form,, that we mentioned before. However, it is not in one of the two indeterminate form that we an apply L Hospital s rule to. In order to turn this into a form that we an apply L Hospital s rule, we use the fat that for any real number x, x = 1 x = 1 1 1/x In order to apply L Hospital s Rule, we first rewrite x ln x = ln x 1/x As x +, ln x and 1, now we may apply L Hospital s Rule: x
ln x x ln x = x + x + 1/x = 1/x x + 1/x = 2 x x2 + x = x = x + Evaluate the it: ( 3x 2 + 1 ) 1/x 2 Ans: This expression is another indeterminate form:. L Hospital s rule on this, we use the fat that: In order to apply ln ( a b) = b ln a. Let y = ( 3x 2 + 1 ) 1/x 2, then ln(y) = ln [ (3x 2 + 1 ) 1/x 2] = 1 x 2 ln ( 3x 2 + 1 ). 1 We first evaluate: ln(y) = x ln ( 3x 2 + 1 ) 2 1 x ln ( 3x 2 + 1 ) ln ( 3x 2 + 1 ) = 2 x 2 As x, ln(3x 2 + 1), and x 2, so we may apply L Hospital s rule: ln ( 3x 2 + 1 ) 6x/(3x 2 + 1) 6x = x 2 = 2x 2x (3x 2 + 1) = 6x 6x 3 + 2x We apply L Hospital s rule one more time: 6x 6x 3 + 2x = So we have: [ 1 ln ( 3x 2 + 1 ) x 2 ] 6 18x 2 + 2 = = ln y = Using the fat that ln x is a ontinuous funtion, we have: ( ln(y) = ln y ) = y = e = 1 ( 3x 2 + 1 ) 1/x 2 = 1 Understand that L Hospital s rule usually does not allow you to find the it of a funtion diretly. Instead, it allows you to hange an indeterminate form to a format that is no longer indeterminate, at whih point you may be able to find the value of the it using other knowledge. 2x + 3 Evaluate: e x Answer: As x, both 2x + 3 and e x, so this is an indeterminate
form, we may apply L Hospital s Rule: 2x + 3 e x 2 = e x At this point, the it is no longer indeterminate, sine the numerator does not approah as x. We may no longer use L Hospital s rule, but we do not need to. As x, the denominator e x while the numerator is a onstant, this is the finite divided by infinite ase, the result is : 2 e x =