# Did I prove this limit correctly?

Given $\lim_{n \to \infty}a_n = L$ and $\lim_{n \to \infty}b_n = M$ implies that $\lim_{n \to \infty}2a_n + 3b_n = 2L + 3M$

Proof

Assume $\lim_{n \to \infty}a_n = L$ and $\lim_{n \to \infty}b_n = M$ and $\forall \epsilon > 0$, we have $\frac{\epsilon}{4}>0$ and $\frac{\epsilon}{6}>0$

$|a_n - L| < \frac{\epsilon}{4}$ when $n > N_1$

$|b_n - M| < \frac{\epsilon}{6}$ when $n > N_2$

Set $n>N= \max \left \{ N_1, N_2 \right \}$,we have

$|2a_n + 3b_n - (2L + 3M)| \leq |2a_n - 2L| + |3b_n - 3M| = |2||a_n - L| + |3||b_n - M| < 2\frac{\epsilon}{4} + 3\frac{\epsilon}{6} = \epsilon$

Question: I had to work backwards to choose my $\epsilon$, is that a "general" strategy? I omitted my work for finding my $\epsilon$

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Perfect, almost too much so. You worked too hard in working backwards. I would have picked $\frac{\epsilon}{100}$ for both, then we get $\le \frac{2\epsilon}{100}+\frac{3\epsilon}{100}=\frac{5\epsilon}{100}\lt \epsilon$. – André Nicolas Sep 25 '12 at 18:15

Yes, this looks correct. Working backwards to choose $\epsilon$ is also the strategy I'd recommend in general.