Show that the limit of a sequence exists and nonnegative Let $\mathcal{B}(F)$ be the algebra of all operators on a complex Hilbert space $(F,\langle\cdot\;, \;\cdot\rangle)$.

Let $(x_n)$ be a sequence such that $\|x_n\|=1$ and $A,B\in\mathcal{B}(F)$. Assume that
  $$\frac{1}{n}\|B\|-\frac{1}{2n^3} +\frac{1}{2n}\|A\|^2
  +Re\{e^{-i\theta} \langle Bx_n,x_n\rangle \overline{\langle Ax_n,x_n\rangle} \}>0,
$$
  for all $n$.
  I want to prove that
  $$0\leq  \lim_{n\to\infty}Re\{e^{-i\theta} \langle Bx_n,x_n\rangle \overline{\langle Ax_n,x_n\rangle} \}$$

Attempt: By Cauchy-Schwarz we have  $\{\langle Bx_n,x_n\rangle\}$ and $\{\langle Ax_n,x_n\rangle\}$ are bounded. So, we get the result.

Is this correct or we should use subsequences?

 A: It's trivial to see that if the limit in the final inequality exists then this result is true. However, using Cauchy-Schwarz to obtain boundedness only gives you the existence of subsequential limits. 
Unfortunately this is all you can get in general since the limit need not exist. For example consider the complex Hilbert space $\ell^2(\mathbb{C})$, let $\theta = 0$ and $A = \operatorname{Id}$ so that we are interested in the convergence of the sequence $\operatorname{Re} \langle Bx^{(n)}, x^{(n)} \rangle$. Note that for clarity, for $y \in \ell^2(\mathbb{C})$ I reserve the lower index to denote the position in the sequence $y$ so that $y_k \in \mathbb{C}$ whereas $y^{(k)} \in \ell^2(\mathbb{C})$.
Now consider the operator $B: \ell^2 \to \ell^2$ defined by
$$(Bx)_k = (-1)^k x_k.$$
$B$ is an isometry and hence is certainly continuous. However, if $x^{(n)}$ is the complex sequence with $1$ in the $n$-th position and $0$ elsewhere then $$\operatorname{Re}\langle Bx^{(n)}, x^{(n)} \rangle = (-1)^n$$ which does not converge.
