Analysis Limit Question

  • #1

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.
 

Answers and Replies

  • #2
360
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.

Could you prove it for the monomial ##a_m y^m##? Have you learned L'Hôpital's rule yet?
 
  • #3
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?
 
  • #4
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 realised the bottom part of the limit is wrong as after you've differentiated once your gonna 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?
 
  • #5
360
0
Sorry just realised the bottom part of the limit is wrong as after you've differentiated once your gonna 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##.
 

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