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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?

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  • $\begingroup$ These should be the formal definitions of the given limits:$$(\forall \epsilon > 0)(\exists N \in \Bbb{R})(\forall n \in \Bbb{N})(n > N \implies \lvert a_n-L \rvert < \epsilon)$$ $$(\forall \epsilon > 0)(\exists N \in \Bbb{R})(\forall n \in \Bbb{N})(n > N \implies \lvert b_n-M \rvert < \epsilon)$$ $\endgroup$ – Noble Mushtak Apr 18 '16 at 14:45
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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

$$ |(a_{n}+2b_{n})-(L-2M)|=|(a_{n}-L)+(2b_{n}-2M)|\leq|a_{n}-L|+|2b_{n}-2M|<\frac{\epsilon}{2}+\frac{\epsilon}{2}=\epsilon. $$

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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, $$\lim_{n\rightarrow\infty}(a_n+2b_n)=L+2M$$

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