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!
 A: 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.
A: In general, no.
Note we have
$\dfrac{\langle a_n, b_n \rangle}{\Vert a_n \Vert} = \langle \dfrac{a_n}{\Vert a_n \Vert }, b_n \rangle; \tag 1$
since
$\Vert \dfrac{a_n}{\Vert a_n \Vert} \Vert = \dfrac{\Vert a_n \Vert}{\Vert a_n \Vert} = 1, \tag 2$
we have
$e_n = \dfrac{a_n}{\Vert a_n \Vert} \in S^1, \tag 3$
where $S^1$ is the unit circle, is a sequence of unit vectors in $\Bbb R^2$; furthermore, for any sequence of unit vectors $e_n \in S^1$, we can find $a_n \to 0$ such that (3) binds; as a simple example, set
$a_n = \dfrac{e_n}{n}; \tag 4$
then 
$a_n \to 0. \tag 5$
The question then really hinges on finding $\langle e_n, b \rangle$; but if we take
$e_n = \dfrac{(-1)^n b_n}{ \Vert b_n \Vert}, \tag 6$ 
which we can always do once $n$ becomes big enough, since $b_n \to b$ implies $\Vert b_n \Vert \to \Vert b \Vert \ne 0$, 
we have
$\langle e_n, b \rangle =\langle \dfrac{(-1)^n b_n}{ \Vert b_n \Vert}, b \rangle = (-1)^n \langle \dfrac{b_n}{\Vert b_n \Vert}, b \rangle = \dfrac{(-1)^n}{\Vert b_n \Vert} \langle b_n, b \rangle; \tag 7$
but
$\dfrac{1}{\Vert b_n \Vert} \langle b_n, b \rangle \to \dfrac{1}{\Vert b \Vert} \langle b, b \rangle = \dfrac{\Vert b \Vert^2}{\Vert b \Vert} = \Vert b \Vert; \tag 8$
we then have
$\langle e_n, b \rangle \to \Vert b \Vert, \; n \; \text{even}, \tag 9$
and
$\langle e_n, b \rangle \to -\Vert b \Vert, \; n \; \text{odd}, \tag{10}$
so the sequece $\langle e_n, b \rangle$ fails to converge.  Thus the sequence (1) fails to converge as well.
