Using $\epsilon - \delta$ definition of limit, prove $\lim_{x \to a} (3f(x)-5g(x)) = 3\lim_{x \to a} f(x) -5\lim_{x \to a} g(x)$

Question

The $\lim_{x \to a} f(x)$ and $ \lim_{x\to a} g(x)$ exist. Using $\epsilon - \delta$ definition of limit, prove $\lim_{x \to a} (3f(x)-5g(x)) = 3\lim_{x \to a} f(x) -5\lim_{x \to a} g(x)$

I think I know how to solve this, but I have a question about whether I need to include a step… here is my work.

Given:

$$ \forall \epsilon > 0 \space \exists \delta_{\epsilon}’ > 0 \space \forall x\in \mathbb{R} \space 0<|x-a| < \delta_{\epsilon}’ \space |f(x)-L| < \epsilon $$

$$ \forall \epsilon > 0 \space \exists \delta_{\epsilon}’’ > 0 \space \forall x\in \mathbb{R} \space 0<|x-a| < \delta_{\epsilon}’’ \space |g(x) - M| < \epsilon $$

Now for my question. Do I write

$$ \forall \epsilon > 0 \space \exists \delta_{\epsilon} > 0 \space \forall x\in \mathbb{R} \space 0<|x-a| < \delta_{\epsilon} \space |3f(x) - 5g(x) - (3L- 5M)|< \epsilon $$

Or $$ \forall \epsilon > 0 \space \exists \delta_{\epsilon} > 0 \space \forall x\in \mathbb{R} \space 0<|x-a| < \delta_{\epsilon} \space |3f(x) - 5g(x) - (L - M)|< \epsilon $$

If I use the second, then do I first need to show that $lim_{x\to a} 3f(x) = 3lim_{x\to a} f(x)$? If I use the first method then I think I can just choose

$\delta_{\epsilon} = min(\frac{\epsilon}{6}, \frac{\epsilon}{10})$

Answer 1

The former one it should be. The reason is that now you are proving that the limit of a function $3f(x)+5g(x)$ at $a$ is $3L+5M$. If $L,M$ are limits of $f,g,$ respectively at $a$.

The second statement means that the limit of the above function is $L+M$.