Heres the question that I think I have figured out. I was hoping somebody could look over my proof and tell me weather it is valid. Anyway, the question is:

Okay, so heres my attempt at a proof. Please point out any errors you may see or find:

LIMIT PROBLEM

If $\displaystyle \;\; 0 < \epsilon_1 < f(x) \;\;$ for all $\displaystyle \; x$,

and

$\displaystyle \lim_{x \to a} g(x) = 0$

then prove that:

$\displaystyle \lim_{x \to a} \frac{f(x)}{|g(x)|} = \infty$

Now, are there any errors in this proof? And is it always neccesary to mathematically go over every detail in a proof? Like, is my proof to long? Is it okay to make simple assumptions that your thinking isn't neccesary? (Like stating or not stating in the proof that $\displaystyle 0 \leq |a|$ for all $\displaystyle a$) Thanks in advance

MY ATTEMPT AT PROVING THE LIMIT

By the definition of a limit, we know that:

$\displaystyle \lim_{x \to a} g(x) = 0$

Means that for every:

$\displaystyle \epsilon_2 > 0$,theres is a $\displaystyle \delta_1 > 0$ such that:

$\displaystyle 0 < |x - a| < \delta_1$

implies:

$\displaystyle |g(x) - 0| = |g(x)| < \epsilon_2$

The above inequality implies that:

$\displaystyle \frac{1}{\epsilon_2} < \frac{1}{|g(x)|}$

Also, since:

$\displaystyle 0 < \epsilon_1 < f(x)$

Then it follows that (by multiplying $\displaystyle \epsilon_1$ to both sides of the g(x) inequality):

$\displaystyle \frac{\epsilon_1}{\epsilon_2} < \frac{f(x)}{|g(x)|}$

[I don't know if the follwing is neccesary for this proof, but better safe then sorry]

(We need not worry about $\displaystyle \frac{f(x)}{|g(x)|}$ being negitive; since $\displaystyle 0 \leq |a|$ for all $\displaystyle a$, then the denominator is always positive. And since, by the stipulations made by the problem, $\displaystyle 0 < \epsilon_1 < f(x)$ for all $\displaystyle x$ obviously means that $\displaystyle 0 < f(x)$ for all $\displaystyle x$; then it folllows that both the denominator and numerator are positive, therefore $\displaystyle 0 < \frac{f(x)}{|g(x)|}$ for all $\displaystyle x$)

To prove that the limit approaches infinity, we must show that:

$\displaystyle 0 < |x - a| < \delta_2$

Implies that, for any $\displaystyle N$ we have:

$\displaystyle N < \frac{f(x)}{|g(x)|}$.

But we are already there! (atleast I think, as long as there are no errors) From the given limit of g(x), we were able to show that :

$\displaystyle \frac{\epsilon_1}{\epsilon_2} < \frac{f(x)}{|g(x)|}$

So letting:

$\displaystyle N = \frac{\epsilon_1}{\epsilon_2}$

and

$\displaystyle \delta_1 = \delta_2$

and rememebering the stipulations made by the problem of the value of f(x), then the train of implications goes like this:

$\displaystyle 0 < |x - a| < \delta_2$

$\displaystyle \Rightarrow \;\;\; 0 < |x - a| < \delta_1$

$\displaystyle \Rightarrow \;\;\; |g(x) - 0| = |g(x)| < \epsilon_2$

$\displaystyle \Rightarrow \;\;\; \frac{1}{\epsilon_2} < \frac{1}{|g(x)|}$

Remeber that:

$\displaystyle 0 < \epsilon_1 < f(x)$

So:

$\displaystyle [0 < \epsilon_1 < f(x)] \; \wedge \; [\frac{1}{\epsilon_2} < \frac{1}{|g(x)|} ]$

$\displaystyle \Rightarrow \;\;\; \frac{\epsilon_1}{\epsilon_2} < \frac{f(x)}{|g(x)|}$

Which shows that:

$\displaystyle 0 < |x-a| < \delta_2 \;\; \Rightarrow \;\; N < \frac{f(x)}{|g(x)|}$

For:

$\displaystyle N = \frac{\epsilon_1}{\epsilon_2}$

and

$\displaystyle \delta_1 = \delta_2$

FINNALY, this all means that:

$\displaystyle \lim_{x \to a} \frac{f(x)}{|g(x)|} = \infty$

(granted that the stipulations on f(x) are given)