Let $\ell^2$ over $\mathbb{C}$
Let $h \in \ell^2$ such that $\forall n \in \mathbb{N}: h_n \neq 0$ where $h_n$ is nth number in $h$
Let $\{v_m\}_{m \in \mathbb{N}}$ be a sequence of points in $\ell^2$ such that $v_m \to h$ in norm
Let $\{u_m\}_{m \in \mathbb{N}}$ be a sequence of points in $\ell^2$ such that $u_m \to h$ in norm
We denote $v_{m,n}$ the nth number in $v_m$ and $u_{m,n}$ the nth number in $u_m$
I would like to know if it is true that $$ \lim_{m \to \infty} \dfrac{v_{m,m}}{u_{m,m}} =1 $$
Thanks.
Sequence $h = (1,\frac{1}{2},\frac{1}{3},...)$ is in $\ell^2$. We can approach this sequence in $\ell_2$ with both $v_m = h + \frac{1}{m}(1,1,....,1,0,0,...)$ (exactly $m$ first entries are $1$) and similarly $u_m = h + \frac{1}{m^2}(1,1,...,1,0,0,...)$ (similarly here), since $$ \|h-v_m\|_2^2 = \sum_{k=1}^m \frac{1}{m^2} = m \cdot \frac{1}{m^2} =\frac{1}{m} \to 0$$ and $$ \|h-u_m\|_2^2 = \sum_{k=1}^m \frac{1}{m^4} = m \cdot \frac{1}{m^4} = \frac{1}{m^3} \to 0.$$
But, we see that $v_{m,m} = \frac{2}{m}$, $u_{m,m} = \frac{1}{m} + \frac{1}{m^2}$ giving us $\frac{v_{m,m}}{u_{m,m}} = \frac{2}{1+\frac{1}{m}} \to 2$. You can easily modify above to get any value as a limit