Absolute value manipulation in limit proof

Question

Given $|a_n - L| < \epsilon$ for $n > N$ and $|b_n - a_n| < \epsilon$ for $n > M$ we can conclude $|b_n - L| < k\epsilon$ for $n > max(N, M)$ and some $k$ (e.g. $2$). I understand this intuitively, but how can I show it by algebraic manipulation? I suppose I should use the triangle inequality but I cannot figure out exactly how.

Answer 1

$|b_n - L|=|(b_n - a_n)+(a_n- L)|\leq |b_n - a_n|+|a_n- L|<2\epsilon$ for $n>\max (N,M)$, using as you suggested the triangle inequality.

Answer 2

If you choose $n>\max\{N,M\}$ than

$|b_n-L|=|(b_n-a_n)+(a_n-L)|<$

$<|(b_n-a_n)|+|(a_n-L)|<\epsilon+\epsilon=2\epsilon$