I have the following quotient $$ \frac{\int_k^\infty x^1 \ f(x) dx}{\int_k^\infty x^{1+\delta} \ f(x) dx} $$ Where $f(x)$ is bounded and $\delta>0$.
I am trying to prove that as $k\to\infty$ the quotient becomes 0. If it were necessary, I can also assume that both integrals exist and are bounded, but it doesn't seem necessary. Also, $f(x)$ is a pdf, and it would be ok to show the statement for continuous functions.
It seems to be pretty intuitive but I cannot come up with an formal way of showing it. I am just not used to dealing with quotients of integrals.
Let $(a_k,b_k)=\left(\int_k^\infty xf(x)\,dx, \int_k^\infty x^{1+\delta}f(x)\,dx\right)$ both be finite. The integrand in the $b_k$ integral is pointwise greater than $x k^\delta f(x)$, so $b_k\ge k^\delta a_k$, and so $a_k/b_k\le k^{-\delta}$, which tends to $0$ as $k\to \infty,$ as desired.