I'm hoping someone might have a hint as to how to prove/disprove the follow claim:
Claim: Let $f,g$ be real valued functions. If $\lim_{x\rightarrow\infty}f(x)g(x)\text{ exists}\space\text{and}\space\lim_{x\rightarrow\infty}f(x)=0,$ then $\lim_{x\rightarrow\infty}f(x)g(x)=0.$
Broader context: I'm trying to show that if $\mathbb{P}\left(X\leqslant0\right)=1\space\text{and}\space\mathbb{E}\left[|X|^{n}\right]<\infty,$ then for $n>0, $ $\mathbb{E}\left[X^{n}\right]=-n\int_{-\infty}^{0}x^{n-1}F_{x}(x)dx.$
I get stuck on the following step: $\mathbb{E}\left[X^{n}\right] =\lim_{L\rightarrow\infty}\left(\left[x^{n}F_{x}(x)\right]_{-L}^{0}-n\int_{-L}^{0}x^{n-1}F_{x}(x)dx\right) =\cdots.$ It's clear from the question that $\lim_{L\rightarrow\infty}\left[x^{n}F_{x}(x)\right]_{-L}^{0}=0$ but I'm not clear on what makes $\lim_{L\rightarrow\infty}\left[-(-L)^{n}F_{x}(-L)\right]$ evaluable.
Your claim is not true. Consider $f(x) = \frac{1}{x}$ and $g(x) = x$. Then $\lim\limits_{x \to \infty} f(x) = 0$ but $\lim\limits_{x \to \infty} f(x) g(x) = 1$.
Now, let's solve your problem.
We know that $F_x(L)$ is defined to be $\mathbb{P}(x \leq L)$. We must show that $\lim\limits_{L \to \infty} (-L)^n \mathbb{P}(x \leq -L) = 0$.
We see that $\mathbb{E}(|X|^n$ given $X \leq -L) \mathbb{P}(x \leq -L) + \mathbb{E}(|X|^n$ given $X > -L) \mathbb{P}(X > -L) = \mathbb{E}[|X|^n]$. Then $\mathbb{E}(|X|^n$ given $X \leq -L) \mathbb{P}(x \leq -L) = \mathbb{E}[|X|^n] - \mathbb{E}(|X|^n$ given $X > -L) \mathbb{P}(X > -L)$.
Taking the limit as $L$ goes to infinity gives us $\lim\limits_{L \to \infty} \mathbb{E}(|X|^n$ given $X \leq -L) \mathbb{P}(x \leq -L) = 0$.
And we see that $L^n \leq \mathbb{E}[|X|^n$ given $X \leq -L]$. So therefore, $\lim\limits_{L \to \infty} L^n P(X \leq -L) = 0$.