Prove that $\langle Ax_n\;,\;x_n\rangle\to\langle Ax\;,\;x\rangle$ In this question $F$ stands for a complex Hilbert space with inner product $\langle\cdot\;,\;\cdot\rangle$ and the norm $\|\cdot\|$. Let $\mathcal{B}(F)$ be the algebra of all bounded linear operators on $F$.

Assume that there exists $(x_n)_n\subset F$ such that $\lim_{n\to \infty}\|x_n-x\|=0$. If $A\in\mathcal{B}(F)$, then why is the following true:
  $$\lim_{n\to \infty}\langle Ax_n\;,\;x_n\rangle=\langle Ax\;,\;x\rangle ?$$

 A: Write 
$$
\begin{align}
|\langle Ax_n, x_n  \rangle-\langle Ax, x  \rangle|&\leq |\langle A(x_n-x), x_n  \rangle|+ |\langle Ax, x_n-x  \rangle|\\
&\leq \lVert A\rVert\lVert x_n-x\rVert\lVert x_n\rVert +  \lVert A\rVert\lVert x\rVert \lVert x_n-x\rVert
\end{align}
$$
by the C.S inequality. Argue that this can be made less than $\epsilon >0$ by choosing $n$ sufficiently large using the hypotheses. Special care needs to be taken to deal with the cases where $\lVert A\rVert=0$ or $\lVert x\rVert=0$.
A: Consider the map $v\mapsto\langle Av,v\rangle$. It is continuous, since you can get it by composing two continuous maps: the map $v\mapsto(Av,v)$ and the map $(v,w)\mapsto\langle v,w\rangle$. 
Therefore, since $\lim_{n\to\infty}x_n=x$, $\lim_{n\to\infty}\langle Ax_n,x_n\rangle=\langle Ax,x\rangle$.
A: $$| \langle Ax_n,x_n \rangle-\langle Ax,x \rangle| \leq |\langle Ax_n-Ax,x \rangle|+|\langle Ax_n,x_n-x \rangle |$$ $$ \leq \| A\| \| x_n-x\|\|x\|+\|A\|\|x_n\|\|x_n-x\|$$
By the boundedness of $A$ and $\lim_{n\to \infty}\|x_n-x\|=0$ we have the result 
Here you are 
