1.This ain't no joke...I need your help....(adsbygoogle = window.adsbygoogle || []).push({});

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.

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# Homework Help: Antoine L'Hôspital's rule problem

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