What is the solution to the Taylor series limit 0/0?

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SUMMARY

The solution to the limit of the form 0/0 using Taylor series involves identifying the smallest degree factor that dominates as x approaches zero. In this case, the x^4 terms in both the numerator and denominator are critical. By applying L'Hôpital's Rule, one can differentiate the numerator and denominator until reaching a determinate form. The final result of the limit is -1/6, derived from the ratio of the leading x^4 terms.

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Homework Statement



I need to solve this limit in the form 0/0 with the Taylor series...
4ecd4838e4b04e045aee647b-alfie-1322076258534-wolframalpha20111123132253954.gif


Homework Equations





The Attempt at a Solution


Alright, I didn't really get where I am supposed to "stop" writing polinomials, my teacher said that I should stop when I find the smallest degree factor, because that's the one which is "bossing around" when the limit approaches zero.
Okay, that's where I've gone so far:

http://www4d.wolframalpha.com/Calculate/MSP/MSP619i577b6i0hb357g0000173a0a18c530af8g?MSPStoreType=image/gif&s=64&w=320&h=58

I don't get if I wrote too many, if I didn't write enough terms, if I did something wrong at all, or I am right and should keep on doing calcs. Could someone help me out please? Thanks.
 
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Collect like terms in the numerator. What does that give you?

And what about the denominator?
 
-(2 x^4)/3-(119 x^6)/120-(1261 x^8)/5040
OVER
x^6+4x^5+4x^4

Even if I collect like terms, I get something that doesn't really get me close to the limit, and I think that huge number there is just wrong...
But, I don't get when I have to stop, I mean, I could have gone for infinity keeping on writing the series polynomials. When do I have to stop writing?
 
Remember L’ Hôpital’s Rule. If both f(x) and g(x) approach 0 at some point x0, then to evaluate f(x)/g(x) at x0 you try to evaluate f'(x)/g'(x) at x0. If that still results in the indeterminate form 0/0, you can iterate and try to evaluate f''(x)/g''(x). If that doesn't help, try to evaluate f'''(x)/g'''(x), and so on, until you either reach some form that is not indeterminate or a form that blows up.

Assume that you have a Taylor expansion of f(x) and g(x) about the point of interest\begin{aligned}<br /> f(x) &amp;= \sum_{n=0}^{\infty} a_n (x-x_0)^n \\<br /> g(x) &amp;=\sum_{n=0}^{\infty} b_n (x-x_0)^n<br /> \end{aligned}

where the first few an and bn are zero. (If a0 and b0 are not zero there's no need for this L’ Hôpital rigamarole.)

The limit is
  • Zero if the number of leading zeros in {an} is greater than the number of leading zeros in {bn}.
  • Undefined (infinite) if the number of leading zeros in {an} is less than the number of leading zeros in {bn}.

These cases are kinda uninteresting. This leaves as an interesting case where the number of leading zeros in {an} and {bn} are equal.

Which case applies to your problem?
 
Okay, I tried to follow, the result of my limit is neither 0 nor infinity, so it must be when a_n or b_n are equal... uhm... how can I use such information to help myself into the problem?
 
dttah said:
Okay, I tried to follow, the result of my limit is neither 0 nor infinity, so it must be when a_n or b_n are equal... uhm... how can I use such information to help myself into the problem?

The terms that are "bossing around" (i.e. are dominating as x->0) are the x^4 terms in the numerator and the denominator. Suppose you just look at those. What's the ratio?
 
The terms that are "bossing around" (i.e. are dominating as x->0) are the x^4 terms in the numerator and the denominator. Suppose you just look at those. What's the ratio?

Ahh! Now I get it! :) I just checked with x^4 terms and it pops out -1/6 getting rid of the denominator x^4 term as well... thanks a lot! That's appreciated ! :)
 

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