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Proof

We provide a proof of this theorem in the special case when f , g , f , and g are all continuous over an open interval containing a . In that case, since lim x a f ( x ) = 0 = lim x a g ( x ) and f and g are continuous at a , it follows that f ( a ) = 0 = g ( a ) . Therefore,

lim x a f ( x ) g ( x ) = lim x a f ( x ) f ( a ) g ( x ) g ( a ) since f ( a ) = 0 = g ( a ) = lim x a f ( x ) f ( a ) x a g ( x ) g ( a ) x a algebra = lim x a f ( x ) f ( a ) x a lim x a g ( x ) g ( a ) x a limit of a quotient = f ( a ) g ( a ) definition of the derivative = lim x a f ( x ) lim x a g ( x ) continuity of f and g = lim x a f ( x ) g ( x ) . limit of a quotient

Note that L’Hôpital’s rule states we can calculate the limit of a quotient f g by considering the limit of the quotient of the derivatives f g . It is important to realize that we are not calculating the derivative of the quotient f g .

Applying l’hôpital’s rule (0/0 case)

Evaluate each of the following limits by applying L’Hôpital’s rule.

  1. lim x 0 1 cos x x
  2. lim x 1 sin ( π x ) ln x
  3. lim x e 1 / x 1 1 / x
  4. lim x 0 sin x x x 2
  1. Since the numerator 1 cos x 0 and the denominator x 0 , we can apply L’Hôpital’s rule to evaluate this limit. We have
    lim x 0 1 cos x x = lim x 0 d d x ( 1 cos x ) d d x ( x ) = lim x 0 sin x 1 = lim x 0 ( sin x ) lim x 0 ( 1 ) = 0 1 = 0.
  2. As x 1 , the numerator sin ( π x ) 0 and the denominator ln ( x ) 0 . Therefore, we can apply L’Hôpital’s rule. We obtain
    lim x 1 sin ( π x ) ln x = lim x 1 π cos ( π x ) 1 / x = lim x 1 ( π x ) cos ( π x ) = ( π · 1 ) ( −1 ) = π .
  3. As x , the numerator e 1 / x 1 0 and the denominator ( 1 x ) 0 . Therefore, we can apply L’Hôpital’s rule. We obtain
    lim x e 1 / x 1 1 x = lim x e 1 / x ( −1 x 2 ) ( −1 x 2 ) = lim x e 1 / x = e 0 = 1 · lim x e 1 / x 1 ln x .
  4. As x 0 , both the numerator and denominator approach zero. Therefore, we can apply L’Hôpital’s rule. We obtain
    lim x 0 sin x x x 2 = lim x 0 cos x 1 2 x .

    Since the numerator and denominator of this new quotient both approach zero as x 0 , we apply L’Hôpital’s rule again. In doing so, we see that
    lim x 0 cos x 1 2 x = lim x 0 sin x 2 = 0 .

    Therefore, we conclude that
    lim x 0 sin x x x 2 = 0 .
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Evaluate lim x 0 x tan x .

1

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We can also use L’Hôpital’s rule to evaluate limits of quotients f ( x ) g ( x ) in which f ( x ) ± and g ( x ) ± . Limits of this form are classified as indeterminate forms of type / . Again, note that we are not actually dividing by . Since is not a real number, that is impossible; rather, / . is used to represent a quotient of limits, each of which is or .

L’hôpital’s rule ( / Case)

Suppose f and g are differentiable functions over an open interval containing a , except possibly at a . Suppose lim x a f ( x ) = (or ) and lim x a g ( x ) = (or ) . Then,

lim x a f ( x ) g ( x ) = lim x a f ( x ) g ( x ) ,

assuming the limit on the right exists or is or . This result also holds if the limit is infinite, if a = or , or the limit is one-sided.

Applying l’hôpital’s rule ( / Case)

Evaluate each of the following limits by applying L’Hôpital’s rule.

  1. lim x 3 x + 5 2 x + 1
  2. lim x 0 + ln x cot x
  1. Since 3 x + 5 and 2 x + 1 are first-degree polynomials with positive leading coefficients, lim x ( 3 x + 5 ) = and lim x ( 2 x + 1 ) = . Therefore, we apply L’Hôpital’s rule and obtain
    lim x 3 x + 5 2 x + 1 = lim x 3 + 5 / x 2 x + 1 = lim x 3 2 = 3 2 .

    Note that this limit can also be calculated without invoking L’Hôpital’s rule. Earlier in the chapter we showed how to evaluate such a limit by dividing the numerator and denominator by the highest power of x in the denominator. In doing so, we saw that
    lim x 3 x + 5 2 x + 1 = lim x 3 + 5 / x 2 x + 1 = 3 2 .

    L’Hôpital’s rule provides us with an alternative means of evaluating this type of limit.
  2. Here, lim x 0 + ln x = and lim x 0 + cot x = . Therefore, we can apply L’Hôpital’s rule and obtain
    lim x 0 + ln x cot x = lim x 0 + 1 / x csc 2 x = lim x 0 + 1 x csc 2 x .

    Now as x 0 + , csc 2 x . Therefore, the first term in the denominator is approaching zero and the second term is getting really large. In such a case, anything can happen with the product. Therefore, we cannot make any conclusion yet. To evaluate the limit, we use the definition of csc x to write
    lim x 0 + 1 x csc 2 x = lim x 0 + sin 2 x x .

    Now lim x 0 + sin 2 x = 0 and lim x 0 + x = 0 , so we apply L’Hôpital’s rule again. We find
    lim x 0 + sin 2 x x = lim x 0 + 2 sin x cos x −1 = 0 −1 = 0 .

    We conclude that
    lim x 0 + ln x cot x = 0 .
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Questions & Answers

f(x) =3+8+4
tennesio Reply
d(x)(x)/dx =?
Abdul Reply
scope of a curve
Abraham Reply
check continuty at x=1 when f (x)={x^3 if x <1 -4-x^2 if -1 <and= x <and= 10
Raja Reply
what is the value as sinx
Sudam Reply
f (x)=x3_2x+3,a=3
Bilal Reply
given demand function & cost function. x= 6000 - 30p c= 72000 + 60x . . find the break even price & quantities.
Fiseha Reply
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Asif Reply
An airline sells tickets from Tokyo to Detroit for $1200. There are 500 seats available and a typical flight books 350 seats. For every $10 decrease in price, the airline observes and additional 5 seats sold. (a) What should the fare be to maximize profit? (b) How many passeners would be on board?
Ravendra Reply
I would like to know if there exists a second category of integration by substitution
CHIFUNDO Reply
nth differential cofficient of x×x/(x-1)(x-2)
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integral of root of sinx cosx
Wedraj Reply
the number of gallons of water in a tank t minutes after the tank has started to drain is Q(t)=200(30-t)^2.how fast is the water running out at the end of 10 minutes?
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Practice Key Terms 2

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Source:  OpenStax, Calculus volume 1. OpenStax CNX. Feb 05, 2016 Download for free at http://cnx.org/content/col11964/1.2
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