I was wondering how can I show that given any bilinear form $b$ in an Hilbert space $$ \lim_{\mathbf{x}\to 0} \frac{b(\mathbf{x}, \mathbf{x})}{||\mathbf{x}||} = 0 $$ I came up with a few ideas in finite dimensions or with non-negative bilinear forms (Cauchy-Schwarz inequality) but I don't know how to tackle the general case yet I'm pretty sure it's easy.
Thank you in advance
Consider the operator defined on an orthonormal basis $\{ \mathbf{e}_n \}$ by $N \mathbf{e}_n = n \mathbf{e}_n$. $(\mathbf{x} ,N\mathbf{x})$ is a form on the linear span of $\{ \mathbf{e}_n \}$ but if you take $\mathbf{x}_k = \frac{1}{\sqrt k} \mathbf{e}_k$ then $ \lim_{k \to \infty} \Vert \mathbf{x}_k \Vert = 0$ but $(\mathbf{x}_k,N \mathbf{x}_k)=1$.