As I was trying to show the variance of $X$ with the exponential distribution and parameter $\lambda$, I got the following bit,
$[-x^2e^{-\lambda x}]_{0}^{\infty}=\lim_{a\to\infty}[-x^2e^{-\lambda x}]_{0}^{a}$,
which is apparently equal to zero. However, we have that $x^2$ goes to $\infty$ and $e^{\lambda x}$ to 0, as $a\to\infty$.
I tried proving this limit:
Let $(x_n)$ be a sequence with limit $+\infty$. We want to show that for each $\epsilon>0$,
$\exists N\in\mathbb N:n>\mathbb N\implies\lvert x_n^2\cdot e^{-\lambda x_n}\rvert<\epsilon$.
But from here on I'm not sure how to proceed. I can see that $\lvert x_n^2\cdot e^{-\lambda x_n}\rvert=x_n^2\cdot e^{-\lambda x_n}$, but that's all I can think of... Could someone give me a hint?
Answer
hint: First you show $e^{\lambda x_n} > \dfrac{\lambda^3x_n^3}{6}$, but this is evidently true from the expansion Maclaurin of $e^{\lambda x}$, and use Squeeze theorem to settle the result.
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