Bounding e raised to a polynomial - Tell me if this is true?

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The discussion confirms that the statement regarding the bounding of the expression P(x)e^{Q(x)} by Ae^{B|x|^{k+\epsilon}} is true. Here, P(x) is a finite-degree polynomial and Q(x) is a polynomial of integer order k. For any positive epsilon, real numbers A and B can be chosen such that the inequality holds for all x. The intuition provided indicates that the growth of the right-hand side will dominate the left-hand side as x approaches infinity, validating the assertion.

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Skatch
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Can someone tell me if the following statement is true?

Say you have [tex]P(x)e^{Q(x)}[/tex] where P(x) is any finite-degree polynomial and Q(x) is a polynomial of integer order k. Is it true that for any positive epsilon, I can find some real numbers A and B such that

[tex]|P(x)e^{Q(x)}| \leq Ae^{B|x|^{k+\epsilon}}\quad \forall x ?[/tex]

I don't need a proof, just some intuition about why its true or a counterexample if its not. I figure since the expression on the right will be growing faster than that on the left, I should be able to find some A and B to ensure the RHS will always be larger, right? Is this an obvious statement or am I crazy??

Edit: I think its true since the smallest the right hand side can be is A at x=0. So I choose A so that the RHS is larger at x=0, and B is chosen to ensure that its larger around zero, and then far from zero the growth of the exponent on the RHS will dominate and we are home free... This is the idea, yes?
 
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Intuition:
|P(x)| < Ce|x|ε for some C and ε > 0. Similarly |Q(x)| < A|x|k for some A. Put it all together to get what you want.
 
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