Can sombody please "rigoursly" check this proof about a property of limits?

Heres the given problem:

$\emph{Prove \; that \; if: \;}$

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

and:

$|h(x)| \leq M$

then:

$\lim_{x \to 0} g(x) \cdot h(x) = 0$

I am going to put my proof below. I'm going to ask the reader for a big favor; if you happen to see anything at all (and I realy do mean anything at all; even if you just see something that is only the slightest bit incorrect, please still point it out) if you see any mistakes, please nit-pick and split-hairs, and say "well technically thats not...", ect., I think you get the picture. I'd like to be able to one day complete proofs that have the qulaity of near perfect rigor; and I can't do that without some incredibly precise and thorough criticisim (Wait). Heres my attempt at a proof:

Quote:

First, lets make things crystal clear. This is what must be shown to complete the proof:

$\emph{\;For \; every \;} \; \epsilon_2 > 0 \; \emph{\;there \; is \; some \;} \delta_2 > 0 \; <br /> \emph{such \; that, \; for \; all \;} x, \; \emph{we \; have:}$

[1-a]

$\;\;\;\; 0 < |x-0| < \delta_2 \;\;$

[1-b]

$\;\;\;\; \mathrm{([1-a] \; goes \; here) \;}\Rightarrow \; |g(x) \cdot h(x) - 0| < \epsilon_2$

Now, next thing I shall do is to list a few properties that I will be using in my proof, so I need not explain them in the middle of a logical flow, but rather I can referance to them:

[i]

- For any numbers $a$ and $b$ , the follwoing is true:

$|a \cdot b| = |a| \cdot |b|$

[ii]

- For any numbers $\; a, \; b, \; c, \; \mathrm{and} \; d \;$ ; if we know that:

$a < b \; \; \; \mathrm{and} \;\;\; c \leq d$

then the following inequality holds:

$a \cdot c < b \cdot d$

[iii]

- For some mathematical statement $P$ , and some different mathematical statment (which is a function) $Q$ , and arbitrary numbers $x_0$ and $x_1$ (and taking note that the expression denoted by $Q$ is $Q(x)\;\;$ ) ; with all the previous stipulations assumed; and, lastly, knowing the two following statements to be true:

$Q(x_0) \: \Rightarrow \: P$

and that:

$x_0 = x_1$

Then it follows that (I think it follows from the above.... (Worried)) it follows that:

$Q(x_1) \: \Rightarrow \: P$

Now, with those 'properties' out of the way, let me begin the proof. From the given information and the definition of a limit, we know that for every $\epsilon_1 > 0$ there is some $\delta_1 > 0$ such that:

[2]

$\;\; 0 < |x - 0| = |x| < \delta_1 \;\; \Rightarrow \;\; |g(x) - 0| = |g(x)| < \epsilon_1$

The problem also gave us the fact that, for the denoted $M$ ,

[3]

$\;\; h(x) \leq M$

So, combining [2], [3], [i], and [ii] we can conclude that:

[4-a]

$\;\; 0 < |x-0| = |x| < \delta_1$

[4-b]

$\;\; \mathrm{([4-a] \; goes \; here)\;}\Rightarrow \;\; |g(x)| \cdot |h(x)| = |g(x) \cdot h(x)| < \epsilon_1 \cdot M$

Now, as we can see, the above implication between the two inequalities looks allot like what we are trying to prove, [1] that is. Now, we know from [iii] that we can get to [1-a] from [4-a] by simply letting:

$\delta_1 = \delta_2$

And, we can also easily get from [4-b] to [1-b] by letting:

$\epsilon_1 = (\epsilon_2 /M)$

So, from all of the above, we have been able to show from the given information that:

$\emph{\;For \; every \;} \; \epsilon_2 > 0 \; \emph{\;there \; is \; some \;} \delta_2 > 0 \; \emph{such \; that, \; for \; all \;} x, \; \emph{we \; have:}$

[1-a]

$\;\;0 < |x-0| < \delta_2 \;\;$

[1-b]

$\;\; \mathrm{([1-a] \; goes \; here) \;}\Rightarrow \: |g(x) \cdot h(x) - 0| < \epsilon_2$

Which, by the definition of a limit, means that:

$\lim_{x \to 0} \: g(x) \cdot h(x) \: = \: 0$

I think, if I've made no errors above (but, ofcourse, I could have very easily made many errors above, I'm no expert), but if this is correct, I think that completes the proof.

Thank you in advance for any errors or 'technicalities' you may find, and for any advice you can give. Thanks