Epsilon-Delta definition of function limit - Where am I going wrong?

Question

This is an undergrad homework question which I have attempted, and although it has been marked I still cannot spot the flaw in my method. This is my first post so I hope I'm not breaking any rules.

The exact question is:

"Let $c \in \mathbb{R}$ and let $f,g : \mathbb{R} \: \setminus \{ c \} \rightarrow \mathbb{R} $ be real functions.

Suppose that $f(x) \rightarrow L $ and $ g(x) \rightarrow M $ as $ x \rightarrow c.$

Prove directly from the definition that $ \lim_{x \rightarrow c}(4f(x) - 6 g(x)) = 4L - 6M. $"

Here's my attempt:

I appear to be choosing $\delta $ incorrectly. I have a feeling I'm misunderstanding something fundamental.
But I can't see a logical reason why I can't establish the relation between $\epsilon$ and my placeholder $\delta^*$ by taking min(${\delta}_1, \delta_2) $, and then divide through to give me the arbitrary $\epsilon$.

Answer 1

Dividing $\delta^\star$ by $10$ does nothing for you. You have proved that if $0<|x-c|<\delta$ then the conclusion $|BLAH| < 10\epsilon$ holds. Assuming that $0<|x-c|<\delta/10$ does NOT suddenly strengthen the conclusion to $|BLAH| < \epsilon$. Dividing by $10$ in the domain does not have any kind of effect such as dividing by $10$ in the range.

You would do better to start out your argument by choosing $\delta_1$, $\delta_2$ differently, so that if $0<|x-c|<\delta_1$ then $|f(x)-L| < \epsilon/10$ and if $0<|x-c| < \delta_2$ then $|g(x)-M| < \epsilon/10$. Then everything would work out fine.

Answer 2

Let $\epsilon > 0 $ be given. Since $f(x) \to L $ and $g(x) \to M$ as $x \to c$, we can select $\delta_1, \delta_2 > 0$ so that

$$ |f - L| < \frac{ \epsilon }{4} $$

$$ |g - M | < \frac{ \epsilon }{6} $$

whenever $|x-c| < \delta_{1,2} $

Next, if we put $\delta := \min\big( \delta_1, \delta_2 \big) $, then with $|x-c| <\delta$ we must have:

$$ | 4f + 6g - 4L - 6M | = | 4f - 4L + 6g - 6M | \leq 4|f-L| + 6|g-M| < 4 \frac{ \epsilon}{4} + 6 \frac{ \epsilon}{6} = \epsilon $$

Hence, by definition $4f + 6g \to 4L + 6M $