Let $\mathbf{x}_i \to \mathbf{x} \neq 0$ be a convergent sequence in $\mathbb{R}^n$. Suppose $r_i$ is a sequence of real numbers which may or may not converge, but we are given that $r_i\mathbf{x}_i$ does converge to some $\mathbf{y} \in \mathbb{R}^n$.
Is $\mathbf{y}$ necessarily a scalar multiple of $\mathbf{x}$? I'm almost sure of this, but can't figure out how to prove it.
I'm thinking that the components of $r_i \mathbf{x}_i$ perpendicular to $\mathbf{x}_i$ are zero, so the component of $\mathbf{y}$ perpendicular to $\mathbf{x}$ must also be zero, but I'm not sure how to formally prove this.
No.
For example, imagine $\mathbf{x}_i$ is in $\mathbb{R}$ and $\mathbf{x}_i = \frac{1}{5i+1}$ and $r_i = 10i$ then $y_i \cdot \mathbf{x}_i \to \mathbf{2}$ which is not a multiple of $0$. Also, $r_i \ne \alpha.\mathbf{x}_i$ for any $\alpha \in \mathbb{R}$