Limit of Polynomial over Exponential Equals Zero

  • Thread starter Thread starter tobinator250
  • Start date Start date
  • Tags Tags
    Analysis Limit
Click For Summary

Homework Help Overview

The discussion revolves around demonstrating that the limit of a polynomial \( Q(y) \) divided by an exponential function \( e^{y^2} \) approaches zero as \( y \) approaches infinity. The polynomial is defined as \( Q(y) = a_0 + a_1 y + \ldots + a_m y^m \), where \( m \) is the degree of the polynomial.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the use of L'Hôpital's rule as a potential method to evaluate the limit. There is also mention of proving the statement for a monomial and considering induction as a formal approach. Questions arise regarding the application of differentiation and the conditions under which L'Hôpital's rule can be applied.

Discussion Status

The conversation is ongoing, with participants exploring various methods to approach the limit. Some guidance has been provided regarding the use of L'Hôpital's rule and induction, but no consensus has been reached on a definitive method or solution.

Contextual Notes

Participants note the importance of ensuring that the conditions for applying L'Hôpital's rule are met, specifically the requirement for an indeterminate form. There is also an acknowledgment of the complexity involved in differentiating the expressions involved.

tobinator250
Messages
7
Reaction score
0

Homework Statement


Q(y)=a0+a1y+...+amy^m is a polynomial of degree m and I need to show that:

Lim{y->Inf} Q(y)/ey2=0





Homework Equations





The Attempt at a Solution



It seems obvious but I can't seem to be able to prove it, and don't really know where to start, any help would be much appreciated.
 
Physics news on Phys.org
tobinator250 said:

Homework Statement


Q(y)=a0+a1y+...+amy^m is a polynomial of degree m and I need to show that:

Lim{y->Inf} Q(y)/ey2=0





Homework Equations





The Attempt at a Solution



It seems obvious but I can't seem to be able to prove it, and don't really know where to start, any help would be much appreciated.

Could you prove it for the monomial ##a_m y^m##? Have you learned L'Hôpital's rule yet?
 
Oh ok, didn't think of using l'hospital's rule. So can you just say:

Lim_{y->Inf} Q(y)/ey2 <=> lim_{y->inf} Q(m)(y)/((2y)m).ey2)

<=> Lim_{y->inf} a(m).m!/((2y)m).ey2)=0

Is that right?
 
tobinator250 said:
Oh ok, didn't think of using l'hospital's rule. So can you just say:

Lim_{y->Inf} Q(y)/ey2 <=> lim_{y->inf} Q(m)(y)/((2y)m).ey2)

<=> Lim_{y->inf} a(m).m!/((2y)m).ey2)=0

Is that right?

Sorry just realized the bottom part of the limit is wrong as after you've differentiated once your going to have to use the product rule after that. Could you just use induction on the degree m then, and then use l'hospital's rule to prove it for n+1?
 
tobinator250 said:
Sorry just realized the bottom part of the limit is wrong as after you've differentiated once your going to have to use the product rule after that. Could you just use induction on the degree m then, and then use l'hospital's rule to prove it for n+1?

Sure, you could use induction if you want to be very rigorous. I would probably just be lazy and say, "applying L'Hôpital's rule ##m## times...", but induction is a more formal proof. This sort of argument should work for a monomial ##a_m y^m## or for Q(y) itself. Just make sure the hypotheses of L'Hôpital's rule are satisfied: you need to have an indeterminate form of ##0/0## or ##\infty/\infty##.
 

Similar threads

  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 13 ·
Replies
13
Views
3K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 12 ·
Replies
12
Views
5K
  • · Replies 4 ·
Replies
4
Views
3K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 4 ·
Replies
4
Views
4K
  • · Replies 15 ·
Replies
15
Views
7K
  • · Replies 7 ·
Replies
7
Views
2K