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Antoine L'Hôspital's rule problem

  • #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:

Answers and Replies

  • #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:
  • #3
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i can't see how can you go from (9) to (11), check your algebra..

NM... you changed it already
 
  • #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.
 
  • #5
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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:
  • #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.
 
  • #7
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I think my sight is perfectly fine... I got the numerator = -1 and denomenator = 2/3
what d0 you get?
 
  • #8
dextercioby
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The numerator -3 and the denominator 2/3...

Daniel.
 
  • #9
Gokul43201
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vincentchan said:
I think my sight is perfectly fine... I got the numerator = -1 and denomenator = 2/3
what d0 you get?
Vincent, you're making a sign mistake with the first term. It should be -1 - 2, not 1 - 2.
 
  • #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]
 
  • #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]
 
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