If $\lim_{x\to\infty} a_x = a$, and $\lim_{x\to\infty} b_x = b$, prove a $N - \epsilon$ proof that: $\lim_{x\to\infty} (a_x - 3b_x) = a - 3b.$

Question

If

$\lim_{x\to\infty} a_x = a$, and $\lim_{x\to\infty} b_x = b$, prove a $N - \epsilon$ proof that: $$\lim_{x\to\infty} (a_x - 3b_x) = a - 3b.$$

$a_x = a(x); b_x = b(x)$

---

$\lim_{x\to\infty} a_x = a$ means: $\forall \epsilon>0, \exists N_1>0, x>N_1 \to |a_x - a| < \epsilon$

$\lim_{x\to\infty} b_x = b$ means: $\forall \epsilon>0, \exists N_2>0, x>N_2 \to |b_x - b| < \epsilon$

Picking $N = max (N_1, N_2)$,

$\lim_{x\to\infty} (a_x - 3b_x) = a - 3b$ means: $\forall \epsilon>0, \exists N>0, x>N \to |(a_x - 3b_x) - (a-3b)| < \epsilon$

I can further break this down using the triangle inequality:

$|(a_x - 3b_x) - (a-3b)| \leq |a_x - 3b_x| - |a-3b| < \epsilon$

Now I am stuck. What should be my next step?

Answer 1

If $x > \max(N_1, N_2)$

$|(a_x - 3b_x) - (a-3b)| = |(a_x - a) + (3b - 3b_x)|\le |a_x- a| + 3|b-3b_x| < \epsilon + 3\epsilon = 4 \epsilon$.

Can you tweak that?

Maybe letting $N_1$ be such that $x> N_1 \implies |a_x - a| < \epsilon/4$ and letting $N_2$ be such that $x> N_1 \implies |b_x - b| < \epsilon/4$.

Then if $x > \max(N_1, N_2)$...

Answer 2

Here is how to fix it: You're given an $\epsilon$. Now, choose $N_1$ such that $|a_x-a| < \frac\epsilon4$, and $N_2$ such that $|b_x-b|<\frac\epsilon4$. This makes $$|(a_x-3b_x)-(a-3b)| \leq |a_x-a|+3|b_x - b| < \frac\epsilon4 + \frac{3\epsilon}4 = \epsilon$$All other details in there are exactly as in your attempt.