Antoine L'Hôspital's rule problem

[dextercioby]

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

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

MY SOLUTION:

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

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

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

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

Let's analize the V(x) when $x\rightarrow \pi$.I'm sure u'll find a finite positive number (if I'm not mistaking (that would be smth,i checked 3 times ) is $$3\pi^{2}$$ ),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:
$$\lim_{x\rightarrow \pi} V(x)=V(\pi)=3\pi^{2}=:V$$(12)

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

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

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

Then,by using (15) passim (19),the limit becomes (now u see why i insisted to compute V(x) with "x" equal to $\pi$,because it would have been really messy with "y" ):
$$L=-\frac{V}{\pi} \lim_{y\rightarrow 0}\frac{e^{-\sin y}(e^{2\sin y\cos y+\sin y}-1)}{y(y+2\pi)}$$(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:
$$L=-\frac{V}{\pi} \lim_{y\rightarrow 0}\frac{e^{-\sin y}[e^{\sin y (2\cos y+1)}-1]}{y(y+2\pi)}\cdot \frac{\sin y (2\cos y+1)}{\sin y (2\cos y+1)}$$ (21)

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

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

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

Daniel.

 

[dextercioby]

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

Thenx in advance...

Daniel.

EDIT:It took 10 minutes to "cook" and 1hr & a half to edit...

 

[vincentchan]

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

NM... you changed it already

 

[dextercioby]

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.

 

[vincentchan]

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.)

 

[dextercioby]

Then u need to check your sight,pretty quickly... 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.

 

[vincentchan]

I think my sight is perfectly fine... I got the numerator = -1 and denomenator = 2/3
what d0 you get?

 

[dextercioby]

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

Daniel.

 

[Gokul43201]

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.

 

[Gokul43201]

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

Let, $x = \pi + h$ and write this as the following limit :

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

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

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

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

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

Again neglecting higher order terms ,we get :

$$N = \lim_{h\rightarrow 0}~\sin (\pi + h) - \sin(2\pi + 2h) = -h -2h = -3h$$

So, $$L = N/D = \frac{-3h}{2/3} = -\frac{9}{2}$$

 

[dextercioby]

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

Daniel.

P.S.Pisoiule,de formula lui Bernoulli stii??
$$(1+x)^{n}\sim 1+nx,|x|<<1$$