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

In summary, the student is trying to solve for the limit of a function, and is stuck because they don't understand when to stop collecting terms. The limit is when the number of terms in the Taylor series approaches 0, and the student is able to find this by using L'Hôpital's Rule.
  • #1
dttah
37
0

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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  • #2
Collect like terms in the numerator. What does that give you?

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

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?
 
  • #4
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[tex]\begin{aligned}
f(x) &= \sum_{n=0}^{\infty} a_n (x-x_0)^n \\
g(x) &=\sum_{n=0}^{\infty} b_n (x-x_0)^n
\end{aligned}[/tex]

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?
 
  • #5
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?
 
  • #6
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?
 
  • #7
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 ! :)
 

What is Taylor's series limit 0/0?

Taylor's series limit 0/0 is a mathematical concept that is used to approximate the value of a function at a specific point by using a series of derivatives of the function at that point.

Why is Taylor's series limit 0/0 important?

Taylor's series limit 0/0 is important because it allows us to approximate the value of a function at a specific point, even if the function is not defined at that point. This is useful in many mathematical and scientific applications.

What does it mean when a Taylor's series limit 0/0 is equal to infinity?

When a Taylor's series limit 0/0 is equal to infinity, it means that the function cannot be approximated at that point using Taylor's series. This could be because the function is not continuous or differentiable at that point.

How do you solve for a Taylor's series limit 0/0?

To solve for a Taylor's series limit 0/0, you must first find the derivatives of the function at the specific point. Then, you can use the formula for Taylor's series to find the coefficients of the series. Finally, you can use the series to approximate the value of the function at the given point.

What are the applications of Taylor's series limit 0/0?

Taylor's series limit 0/0 has many applications in mathematics and science. It is commonly used in calculus for approximating values of functions, in physics for modeling physical systems, and in engineering for designing systems and structures.

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