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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) = x-2 g(x) = 3x + 5 fog(x)? f(x)/g(x)
Naufal Reply
fog(x)= f(g(x)) = x-2 = 3x+5-2 = 3x+3 f(x)/g(x)= x-2/3x+5
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(mathematics) For a complex number a+bi, the principal square root of the sum of the squares of its real and imaginary parts, √a2+b2 . Denoted by | |. The absolute value |x| of a real number x is √x2 , which is equal to x if x is non-negative, and −x if x is negative.
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log tan (x/4+x/2)
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f(x)=1/1+x^2 |=[-3,1]
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You have downloaded the aplication Calculus Volume 1, tackling about lessons for (mostly) college freshmen, Calculus 1: Differential, and this group I think aims to let concerns and questions from students who want to clarify something about the subject. Well, this is what I guess so.
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CALCULUS
v=(x,y) وu=(x,y ) ∂u/∂x* ∂x/∂u +∂v/∂x*∂x/∂v=1
log tan (x/4+x/2)
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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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