How to find $\lim_{ \| {\bf x} \| \to 0} \frac{ \left( <{\bf x},{\bf y}> \right)^2}{\| {\bf x} \|^2 }$

Question

Let ${\bf x}$ be a vector in $\mathbb{R}^n$, such that $\|{\bf x}\|=r$.

I want to find the following limit \begin{align} \lim_{r \to 0} \frac{ \left(Tr({\bf x} {\bf y}^T) \right)^2}{r^2} \end{align}

for some ${\bf y} \in \mathbb{R}^n$.

Or in other words, by using the language of inner products and norms \begin{align} \lim_{ \| {\bf x} \| \to 0} \frac{ \left( <{\bf x},{\bf y}> \right)^2}{\| {\bf x} \|^2 } \end{align}

By Cauchy-Scwart, we can show that

\begin{align} \lim_{ \| {\bf x} \| \to 0} \frac{ \left( <{\bf x},{\bf y}> \right)^2}{\| {\bf x} \|^2 } \le \lim_{ \| {\bf x} \| \to 0} \frac{ \|{\bf x}\|^2 \|{\bf y}\|^2}{\|{\bf x}\|^2} \le \|{\bf y}\|^2 \end{align}

Note, that for the case of $n=1$ \begin{align} \lim_{x \to 0}\frac{(xy)^2}{x^2}=y^2 \end{align}

Answer 1

The limit doesn't exist. You could pick $x$ with $\|x\| = r$ so that $\langle x,y \rangle = 0$, or $\langle x,y \rangle = r\|y\|$, or anything in between. As $r \to 0$ you could get any value in between $0$ and $\|y\|^2$.