Limit of Polynomial over Exponential Equals Zero

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SUMMARY

The limit of a polynomial Q(y) of degree m divided by e^(y^2) approaches zero as y approaches infinity, formally expressed as Lim{y->Inf} Q(y)/e^(y^2) = 0. This conclusion can be rigorously proven using L'Hôpital's rule, which requires the limit to be in an indeterminate form of 0/0 or ∞/∞. An alternative proof method involves mathematical induction on the degree of the polynomial. Both approaches confirm that the exponential function grows significantly faster than any polynomial function as y increases.

PREREQUISITES
  • Understanding of polynomial functions and their degrees.
  • Familiarity with L'Hôpital's rule for evaluating limits.
  • Basic knowledge of exponential functions, specifically e^(y^2).
  • Concept of mathematical induction for formal proofs.
NEXT STEPS
  • Study the application of L'Hôpital's rule in various limit problems.
  • Explore mathematical induction techniques in proofs involving polynomials.
  • Learn about the growth rates of functions, particularly comparing polynomials and exponentials.
  • Review examples of limits involving indeterminate forms to solidify understanding.
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Students studying calculus, mathematicians interested in limits and growth rates, and educators teaching advanced mathematical concepts related to polynomials and exponentials.

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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.
 
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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##.
 

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