Limit problems.. 
http://s23.postimg.org/xedyol4kr/limit.jpg
I got stuck on Q2 , could someone shed light on how to do it? The proof bit. Finding limits in terms of α and β is easy.
As for Q3, I know it is probably not the way I was supposed to do but is the following method not correct?
Lim (A+B)= lim(a)+lim(b) and we are given that lim(a)=∞ and lim(b) is bounded below hence it is converging to some numerical value L. So lim(a)+lim(b)= ∞+L=∞ ? 
Many thanks
 A: Take $a,b\in \Bbb R$ and let $x=a-b$ and $y=a+3b$. Can you express $a$ and $b$ in terms of $x$ and $y$? If you can, then you can use the exact same thing for the sequences and then use the fact that a linear combination of sequence converges to the linear combination of the limits.
For $3$, no, your method doesn't work because it supposes the convergence of $b_n$ while taking $b_n=\left(-1\right)^n$ is bounded below and doesn't converge. For this question, you should just use the definition of convergence to $+\infty$. Intuitively, since you can get as big as you want, if you get pulled back by something at most finite, then you can still get as big as you want. Now all you have to do it write that with inequalities and quantifiers.
A: I am reading that you have 2a, but want 2b.
Let $a_n=(-1)^n$, and $b_n=-(-1)^n$. Then their sum converges, as
$$a_n+b_n=(-1)^n +-(-1)^n =0.$$
Their product also converges, as
$$a_nb_n=(-1)^n\cdot -(-1)^n=-(-1)^{2n}=-1 \, \forall \, n \in \mathbb{N}.$$
But neither $a_n$, nor $b_n$ converge. They just oscillate perpetually between -1 and 1.
A: Pure arithmetic of limits:
$$2(a_n-b_n)+(a_n+3b_n)=3a_n+b_n \;\;\text{converges}\implies$$
$$(3a_n+b_n)-3(a_n+3b_n)=-6b_n\;\;\text{converges}\implies b_n=-\frac16(-6b_n)\;\;\text{converges}$$
Try a similar trick now to show $\;\{a_n\}\;$ converges
