- #1
JG89
- 728
- 1
Homework Statement
Prove that [tex] \lim_{x \rightarrow \infty} exp(-x^2) \int_0^x exp(t^2) dt = 0 [/tex].
Homework Equations
The Attempt at a Solution
This question is giving me a lot of difficulty. I've tried a lot of different ways to do it, here is a list of ways that I've tried.
1) For t>= 1, we have [tex] 0 < exp(-x^2) \int_0^x exp(t^2) dt \le exp(-x^2) \int_0^x t e^(t^2) dt [/tex]. Letting x tend to infinity, the right side of the inequality tends to 1/2 and so the expression which we wish to find the limit of is bounded and monotonically decreasing, thus it must converge.
2) I've tried approximating e^(t^2) by (1 + t^2/n)^n for large enough n, and then expanding using the binomial theorem and integrating term by term, giving me a polynomial of degree n + 1. We can then find a number m such that x^m > the polynomial in question for large enough x. Then we must find the limit of x^m/e^(x^2) as x tends to infinity. Using the theorem that e^(x^2) becomes infinite of a lower order of magnitude than x^m, we know that this quotient must tend to 0. I'm a bit sketchy about this one because I think m must tend to infinity with x, which complicates things. I think it's best to drop this idea.
3) I figured that the integral of t^(1/n) e^(t^2) for t>=1 decreases monotonically towards e^(t^2) for increasing n, and so if I can show that this integral divided by e^(x^2) tends to 0 for increasing x, then that proves what I want to prove. I've tried integrating it with Mathematica but it starts talking about hypergeometric functions, which I know nothing about so I don't think this is a good approach at all.
Any ideas?