How to prove $\lim\limits_{n\to \infty}(a_n+2b_n)=L+2M$ using the formal definition of limit of sequence? Let $\{a_n\}^{\infty}_{n=1},\{b_n\}^{\infty}_{n=1}$ be sequence. Assume $\lim\limits_{n\to \infty}a_n=L,\ \ \lim\limits_{n\to \infty}b_n=M$, how to prove $\lim\limits_{n\to \infty}(a_n+2b_n)=L+2M$ using the formal definition of limit of sequence? 
I remember the formal definition goes as following(not sure for sequence though): $\forall \epsilon\gt 0,\exists N\in\Bbb{R}:x\gt N\implies |a_n-L|\lt \epsilon$ 
Then we can set $\forall \epsilon\gt 0,\exists N\in\Bbb{R}:x\gt N\implies |a_n-L|\lt \frac12\epsilon$ 
$\forall \epsilon\gt 0,\exists N\in\Bbb{R}:x\gt N\implies |b_n-M|\lt \frac14\epsilon$ 
Then $\forall \epsilon\gt 0,\exists N\in\Bbb{R}:x\gt N\implies |2b_n-2M|\lt \frac12\epsilon$ 
Then $|a_n-L|+|2b_n-2M|\lt \epsilon$.
But I am not sure write a proof, could someone help?
 A: You pretty much got all of the ingredients you need for a proof. Here is how I would put it together. 
Let $\epsilon>0$ be arbitrary.
Since $\lim_{n\to\infty}a_{n}=L$, we may choose an $N_{1}$ such that $|a_{n}-L|<\epsilon/2$.
Since $\lim_{n\to\infty}b_{n}=M$, we may choose an $N_{2}$ such that $|b_{n}-M|<\epsilon/4\implies |2b_{m}-2M|<\epsilon/2$.
Let $N=\max\{N_{1},N_{2}\}$. Then, if $n\geq N$, we have
$$
|(a_{n}+2b_{n})-(L-2M)|=|(a_{n}-L)+(2b_{n}-2M)|\leq|a_{n}-L|+|2b_{n}-2M|<\frac{\epsilon}{2}+\frac{\epsilon}{2}=\epsilon.
$$
A: Let $\epsilon\gt0$ be given. Then, there exist $N_1,N_2\in\mathbb N$ such that
$$|a_n-L|\lt\epsilon$$
for all $n\ge N_1$
and
$$|b_n-M|\lt\frac{\epsilon}2$$
for all $n\ge N_2$
Thus,
$$|2b_n-2M|\lt{\epsilon}$$
for all $n\ge N_2$
Adding the two inequalities,
$$|a_n-L|+|2b_n-2M|\lt2\epsilon$$
for all $n\ge\max{(N_1,N_2)}$
By the triangle inequality,
$$|a_n+2b_n-L-2M|\le|a_n-L|+|2b_n-2M|\lt2\epsilon$$
for all $n\ge\max{(N_1,N_2)}$
Here, $2\epsilon$, like $\epsilon$, is arbitrarily small.
Thus,
$$\lim_{n\rightarrow\infty}(a_n+2b_n)=L+2M$$
