Does $\frac{\langle a_k,b_k\rangle}{\|a_k\|}$ converge, if $(a_k)_k$ and $(b_k)_k$ tend to $0$ and $b\neq 0$ respectively? If $(a_k)_k$ and $(b_k)_k$ both convergent sequences in $\mathbb{R}^2$ such that their limits are $0$ and $b\neq0$ respectively. Does the sequence.
$$\frac{\langle a_k,b_k \rangle}{\|a_k\|}$$ converge? (where $\langle a_k,b_k \rangle$ is the scalar product).
I'm not really sure how to go about this, since the numerator and the denominator both go to $0$( all norms are equivalent so it doesn't matter which norm we choose for the denominator).
I figured I might try using the sandwich theorem to approximate this but I'm not sure how, since the numerator is basically $a^1_kb^1_k * a^2_kb^2_k$ and if $(a_k)_k$ goes to $(0,0)$ obviously the component sequences both tend to $0$ and i can't separate them from the component sequences of $b_k$. So I'm not sure how to go about this!
Any hints at all would be appreciated
Thanks in advance!

The answer is no.
Let $$a_n = \left((-1)^n\cdot\frac1n, \frac1n\right), \text{ for } n\in \mathbb{N}$$
and $b_n = (1, 1)$ for $n \in \mathbb{N}$, the constant sequence $(1,1)$.
We have:
$$\frac{\langle a_n, b_n\rangle}{\|a_n\|} = \frac{(-1)^n\cdot\frac1n + \frac1n}{\frac{\sqrt{2}}{n}} = \frac{(-1)^n + 1}{\sqrt{2}}$$
This sequence does not converge.
