1
$\begingroup$

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?

  • 0
    Possible hint among many: write it as $x^2 / e^{\lambda x}$ and use l'Hopital. – 2017-01-23
  • 0
    We've never properly had l'Hopital in class, even though it's seems to come in handy so often:( – 2017-01-23

1 Answers 1

3

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.