Let $A: H\to H$ and $B: H\to H$ be two compact operators defined on a Hilbert space $H$. Let $(x_n)$ the sequence of eigenfunctions of $A$.
Since $B$ is compact and $(x_n)$ is a basis for $H$ then $$ \lim_{n\to\infty} \langle B(x_n),x_n\rangle=0 $$ It is possible to prove that $$ \lim_{n\to\infty} \frac{\langle A(x_n),x_n\rangle}{\langle B(x_n),x_n\rangle}=0 $$ ?
Some adds: $A\neq B$, $A$ and $B$ are self-adjoint positive definite, furthermore $\mathcal{R}(A) \bigcap \mathcal{R}(B)= (0)$
No. Look at $H=l^2(\mathbb{N})$ and $x_n=e_n$ are the unit vectors.
Let $A$ be the unique operator defined by $Ae_n = \frac{1}{n} e_{n}$ when $n$ is odd and zero otherwise, let $B$ be the unique operator defined by $Be_n = \frac{1}{n^2}(e_{2n} + e_n)$ when $n$ is odd and $B_{e_n} = \frac{1}{n^2}e_{2n}$ when $n$ is even.
Then for odd $n$'s we have $\frac{\left<Ae_n,e_n\right>}{\left<Be_n,e_n\right>} = n$ and therefore the limit as $n$ goes to infinity does not exists.
On the other hand, the range of $A$ is $l^2(2\mathbb{N}+1)$ (i.e. sequences $a_n\in l^2(\mathbb{N})$ for which $a_{2n}=0$ for all $n$). The image of $l^2(2\mathbb{N})$ under $B$ is $l^2(4\mathbb{N})$ and the image of $l^2(2\mathbb{N}+1)$ under $B$ does not intersect with $l^2(4\mathbb{N})\oplus l^2(2\mathbb{N}+1)$. Therefore the range of $B$ does not intersect with the range of $A$.