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Homework Help: Antoine L'Hôspital's rule problem

  1. Jan 22, 2005 #1

    dextercioby

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    1.This ain't no joke...I need your help....

    2.I came across this
    PROBLEM
    Compute:
    [tex] L=:\lim_{x\rightarrow \pi} \frac{e^{\sin x}-e^{\sin 2x}}{\sqrt[3]{\pi x^{2}}-\pi} [/tex] (1)
    ,without using Antoine L'Hôspital's rule...

    MY SOLUTION:

    I used this identity:
    [tex] A^{3}-B^{3}\equiv (A-B)(A^{2}+AB+B^{2}) [/tex](2)
    ,plugged in these values:
    [tex] A\rightarrow \sqrt[3]{\pi x^{2}} [/tex] (3)
    [tex] B\rightarrow \pi [/tex] (4)

    and ended up with:
    [tex] (\sqrt[3]{\pi x^{2}})^{3}-\pi^{3}=(\sqrt[3]{\pi x^{2}}-\pi)[(\sqrt[3]{\pi x^{2}})^{2}+\pi\sqrt[3]{\pi x^{2}}+\pi^{2}] [/tex] (5)
    .
    Defining the animal from the square paranthesis as:
    [tex] V(x)=:(\sqrt[3]{\pi x^{2}})^{2}+\pi\sqrt[3]{\pi x^{2}}+\pi^{2} [/tex] (6)
    ,and using (6),and the fact that:
    [tex] (\sqrt[3]{\pi x^{2}})^{3}=\pi x^{2} [/tex] (7)
    ,equation (5) becomes:
    [tex] \pi x^{2}-\pi^{3}=(\sqrt[3]{\pi x^{2}}-\pi) V(x) [/tex](8)

    From (8),u can immediately see that:
    [tex] \sqrt[3]{\pi x^{2}}-\pi=\pi\frac{(x-\pi)(x+\pi)}{V(x)} [/tex] (9)
    ,where i made use of the decomposition:
    [tex] x^{2}-\pi^{2}=(x-\pi)(x+\pi) [/tex] (10)
    and the factoring of [itex]\pi [/itex].

    Using (9),the original limit (denoted by me with "L") becomes:
    [tex] L=\frac{1}{\pi} \lim_{x\rightarrow \pi} \frac{e^{\sin x}-e^{\sin 2x}}{(x-\pi)(x+\pi)} V(x) [/tex] (11)

    Let's analize the V(x) when [itex] x\rightarrow \pi [/itex].I'm sure u'll find a finite positive number (if i'm not mistaking (that would be smth,i checked 3 times :tongue2: ) is [tex] 3\pi^{2} [/tex] ),anyway,it's relevant that it doesn't really affect the limit in what comes next (of course,the obvious factoring and change of variable);i'll denote it simply by "V",videlicet:
    [tex] \lim_{x\rightarrow \pi} V(x)=V(\pi)=3\pi^{2}=:V [/tex](12)

    Then using (12) and the famous identity:
    [tex] \sin 2x\equiv 2\sin x \cos x [/tex] (13)
    and factoring a minus sign and the exp(onential) with the smaller argument,i get:
    [tex] L=-\frac{V}{\pi}\lim_{x\rightarrow \pi} \frac{e^{\sin x}(e^{2\sin x\cos x-\sin x}-1)}{(x-\pi)(x+\pi)} [/tex] (14)

    Now comes the "tricky part".The substitution
    [tex] x\rightarrow y+\pi [/tex] (15)
    ,under which:
    [tex] \lim_{x\rightarrow \pi}\rightarrow \lim_{y\rightarrow 0} [/tex](16)

    [tex] \sin x\rightarrow \sin (y+\pi)=-\sin y [/tex] (17)
    [tex] \cos x\rightarrow \cos(y+\pi)=-\cos y [/tex] (18)
    and the denominator:
    [tex] (x-\pi)(x+\pi)\rightarrow y(y+2\pi) [/tex] (19)

    Then,by using (15) passim (19),the limit becomes (now u see why i insisted to compute V(x) with "x" equal to [itex] \pi [/itex],because it would have been really messy with "y" :tongue2: ):
    [tex] L=-\frac{V}{\pi} \lim_{y\rightarrow 0}\frac{e^{-\sin y}(e^{2\sin y\cos y+\sin y}-1)}{y(y+2\pi)} [/tex](20)

    And i now i pull the rabbit out of the hat:
    I multiply and divide under the limit with the factor appearing in the exponential in the round bracket:
    [tex] L=-\frac{V}{\pi} \lim_{y\rightarrow 0}\frac{e^{-\sin y}[e^{\sin y (2\cos y+1)}-1]}{y(y+2\pi)}\cdot \frac{\sin y (2\cos y+1)}{\sin y (2\cos y+1)} [/tex] (21)

    and make use SYMULTANEOUSLY of the famous limit:
    [tex] \lim_{y\rightarrow 0} \frac{\sin y}{y} =1 [/tex] (22)
    and of another trick:U know that if [itex] y\rightarrow 0 [/itex] (under the limit,of course),then the whole product [itex] \sin y (2\cos y+1) [/itex]
    goes to zero as well,trick which allows me to make use of another famous limit
    [tex] \lim_{y \rightarrow 0} \frac{e^{y}-1}{y} =\ln e=1 [/tex] (23)

    Then i swich numerators and denominators in (21) as to make use of the relations (22) and (23),to finally get the result
    [tex] L=-\frac{V}{\pi} \frac{3}{2\pi} [/tex] (24)
    and then i make use of the relation (12) which gives me V,i finally obtain
    [tex] L=-\frac{9}{2} [/tex] (25)

    Hopefully i didn't f*** up any calculations... :uhh:

    Daniel.
     
    Last edited: Jan 22, 2005
  2. jcsd
  3. Jan 22, 2005 #2

    dextercioby

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    Sorry,i forgot,i need your help for a simpler solution...

    Thenx in advance... o:)

    Daniel.

    EDIT:It took 10 minutes to "cook" and 1hr & a half to edit... :tongue2:
     
    Last edited: Jan 22, 2005
  4. Jan 22, 2005 #3
    i can't see how can you go from (9) to (11), check your algebra..

    NM... you changed it already
     
  5. Jan 22, 2005 #4

    dextercioby

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    It's okay,trust me...Check it more carefully...The denominator becomes pi times the product in the squares and the V function "goes upstairs" (it would have been at the denominator of the denominator,hence at the numerator).

    Daniel.
     
  6. Jan 22, 2005 #5
    One very simple question, why don't you apply L'hospital rule at the first place?

    edit: the answer should be -3/2 (i didn't use pen and paper, just eyeball the answer.)
     
    Last edited: Jan 22, 2005
  7. Jan 22, 2005 #6

    dextercioby

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    Then u need to check your sight,pretty quickly... :tongue2: L'Hôspital's rule yields the same answer...

    Because my sweet girlfriend doesn't know how to differentiate and to apply L'Hôspital's rule???

    Daniel.
     
  8. Jan 22, 2005 #7
    I think my sight is perfectly fine... I got the numerator = -1 and denomenator = 2/3
    what d0 you get?
     
  9. Jan 22, 2005 #8

    dextercioby

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    The numerator -3 and the denominator 2/3...

    Daniel.
     
  10. Jan 23, 2005 #9

    Gokul43201

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    Vincent, you're making a sign mistake with the first term. It should be -1 - 2, not 1 - 2.
     
  11. Jan 23, 2005 #10

    Gokul43201

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    [tex] L=\lim_{x\rightarrow \pi} \frac{e^{\sin x}-e^{\sin 2x}}{\sqrt[3]{\pi x^{2}}-\pi} [/tex]

    Let, [itex] x = \pi + h [/itex] and write this as the following limit :

    [tex]L =\lim_{h\rightarrow 0} \frac{e^{\sin (\pi + h)}-e^{\sin(2\pi + 2h)}}{\sqrt[3]{\pi (\pi + h)^{2}}-\pi} = N/D~, ~~say[/tex]

    [tex] D = \lim_{h\rightarrow 0}~(\pi ^3 + 2\pi ^2h + \pi h^2)^{1/3} - \pi [/tex]

    Expanding and throwing away second (and higher) order terms in h, you get

    [tex] D = \lim_{h\rightarrow 0}~\pi (1 + \frac {2h}{\pi})^{1/3} - \pi = \frac {2h}{3} [/tex]

    [tex] N = \lim_{h\rightarrow 0}~e^{\sin (\pi + h)}-e^{\sin(2\pi + 2h}) = 1 + \sin (\pi + h) + ... - 1 - \sin(2\pi + 2h) - ... [/tex]

    Again neglecting higher order terms ,we get :

    [tex] N = \lim_{h\rightarrow 0}~\sin (\pi + h) - \sin(2\pi + 2h) = -h -2h = -3h [/tex]

    So, [tex] L = N/D = \frac{-3h}{2/3} = -\frac{9}{2}[/tex]
     
  12. Jan 23, 2005 #11

    dextercioby

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    Thank You for the effort,Gokul,i knew nobody could come up with a simpler version than mine... :tongue2: Yours is more complicated than using L'Hôspital...Yet,u have more points than Vincentchan...He should really check his eyesight... :tongue2:

    Daniel.

    P.S.Pisoiule,de formula lui Bernoulli stii??
    [tex] (1+x)^{n}\sim 1+nx,|x|<<1 [/tex]
     
    Last edited: Jan 23, 2005
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