Limit of Polynomial over Exponential Equals Zero

  • Thread starter Thread starter tobinator250
  • Start date Start date
  • Tags Tags
    Analysis Limit
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'Hopitals 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'Hopitals 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'hopitals 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'hopitals 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##.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
Back
Top