# Antoine L'Hôspital's rule problem

Homework Helper
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 :tongue2: ) 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" :tongue2: ):
$$L=-\frac{V}{\pi} \lim_{y\rightarrow 0}\frac{e^{-\sin y}(e^{2\sin y\cos y+\sin y}-1)}{y(y+2\pi)}$$(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... :uhh:

Daniel.

Last edited:

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

Daniel.

EDIT:It took 10 minutes to "cook" and 1hr & a half to edit... :tongue2:

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

Homework Helper
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.

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:
Homework Helper
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.

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

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

Daniel.

Gokul43201
Staff Emeritus
Gold Member
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.

Gokul43201
Staff Emeritus
Gold Member
$$L=\lim_{x\rightarrow \pi} \frac{e^{\sin x}-e^{\sin 2x}}{\sqrt[3]{\pi x^{2}}-\pi}$$

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

$$(1+x)^{n}\sim 1+nx,|x|<<1$$