# Limit Involving Integral

I have been struggling with the following problem:

$$\lim_{a\to\infty} \int_0^1 \frac {x^2e^x}{(2+ax)} dx .$$

My first reaction was to attempt to move the integral outside the limit, which would make the answer 0. However, I have read a little about Fatou's lemma and am not sure if I can do this. Any help would be greatly appreciated. Thank you

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Note that $$0 \leq \dfrac{x^2 e^x}{2+ax} \leq \dfrac{e}{2+ax} \,\,\,\,\, \forall x \in [0,1]$$ Hence, we have $$0 \leq \int_0^1 \dfrac{x^2 e^x}{2+ax} dx \leq \int_0^1 \dfrac{edx}{2+ax} = e \left.\dfrac{\log(2+ax)}a \right \vert_{x=0}^1 = \underbrace{\dfrac{e}a \log \left(\dfrac{2+a}2\right)}_{\to 0 \text{ as } a \to \infty}$$
If you are lazy to evaluate any integrals, then you could also prove using by Lebesgue dominated convergence theorem, by noting the fact that $$f_a(x) = \dfrac{x^2 e^x}{2+ax} \leq \dfrac{x^2 e^x}2 = g(x)$$ and $\displaystyle \int_0^1 \dfrac{x^2e^x}2 dx < \infty$, since the integral is over a compact set and integrand is continuous over this compact set. Hence, we have $$\lim_{a \to \infty} \int_0^1 \dfrac{x^2 e^x}{2+ax} dx = \int_0^1 \lim_{a \to \infty} \dfrac{x^2 e^x}{2+ax} dx = \int_0^1 0dx = 0$$
In general, you are not allowed to get the limit inside the integral. For instance, consider this. $$\lim_{a \to \infty} \int_0^1 \dfrac{dx}{ax}$$ The answer to this $\infty$, whereas if you get the limit inside the integral, the answer is $0$ i.e. $$\lim_{a \to \infty} \int_0^1 \dfrac{dx}{ax} = \infty \neq 0 = \int_0^1 \lim_{a \to \infty} \dfrac{dx}{ax}$$ – user17762 Mar 22 '13 at 4:17