Does limit exist as x approaches zero? Frobenius Method DEQ

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Homework Help Overview

The discussion revolves around finding the limit of the function (4x^2-1)/(4x^2) as x approaches zero, within the context of the Frobenius Method for differential equations.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants explore the application of L'Hôpital's Rule and question whether it can be used in this scenario. There is also discussion about the necessity of limits for non-constant coefficients in the context of the Frobenius Method.

Discussion Status

The discussion is active, with participants sharing their thoughts on the limit and the application of L'Hôpital's Rule. There are differing interpretations regarding the limit's value and the conditions under which L'Hôpital's Rule is applicable.

Contextual Notes

Some participants express confusion about the conditions for applying L'Hôpital's Rule, particularly regarding the behavior of the numerator and denominator as x approaches zero. There is also mention of the need to consider the limits of p(x) and q(x) in the Frobenius Method.

lonewolf219
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Homework Statement



what is the limit of (4x^2-1)/(4x^2)
when x→0

Homework Equations



In order to find the Indicial Equation, do I need to take the limit of p(x) and q(x), the non-constant coefficients? If so, can the limit of this function be found using LH Rule?

The Attempt at a Solution



Please let me know any info you might have about the Frobenius Method, since I am just learning it from my professor's brief notes about it...
 
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lonewolf219 said:

Homework Statement



what is the limit of (4x^2-1)/(4x^2)
when x→0


What do you think? What have you tried?
 
Well, can you use L' Hopital's Rule twice? (8x - 0)/(8x) and then (8/8) = 1 ?

But I'm confused if I need to multiple by x^2 to find q(nault)?

x^2*q(x)=q(nault)

y''(x) + p(x)y'(x) + q(x)y(x) = 0

If so, would the limit be -1/4?

x^2(4x^2-1)/(4x^2) = [(4x^4)/4x^2] - [x^2/(4x^2)] = [x^2] - [1/4] = [x=0] = - 1/4
 
Last edited:
L'Hopital applies only when the numerator and denominator both go to 0. Here, if x= 0, the numerator is -1 but the denominator is 0. Suppose x were some very small number, say x= 0.000001. What would that fraction be? Now, what do you thing the limit is?
 
negative infinity, right?
 

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