For this question, I'm not sure how to complete the proof. Here is what I have so far. Can anyone please help me out?
Let b, c, l, m ∈ R. Let f, g : R → R be functions. Suppose that $\lim_{ x \to \infty} f(x) = l$ and $\lim_{x \to \infty} g(x) = m$ Give a complete and accurate $\epsilon$ − N proof that $\lim_{ x \to \infty}(2bf(x)-3cg(x))$ exists.
$\lim_{ x \to \infty}(2bf(x)-3cg(x)) = 2bl-3cm$
Let $\epsilon > 0$ be arbitrary. Suppose $ n > N_1$ st $|f(x)-l|<\epsilon/2$. Suppose $n > N_2$ st $|g(x)-m| < \epsilon/2$
Pick N = $max(N_1, N_2)$
$|2bf(x)-3cg(x) -(2bl-3cm)|$
$ = |2bf(x)-3cg(x)-2bl+3cm|$
$ = |2bf(x)-2bl| + |3cg(x)-3cm|$
$ = |2b||f(x)-l| + |3c||g(x)-m|< |2b|\frac{\epsilon}{|2b|*2} + |3c|\frac{\epsilon}{|3c|*2} = \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon$