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

• Aug 24th 2010, 03:30 PM
mfetch22
Can sombody please "rigoursly" check this proof about a property of limits?
Heres the given problem:

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

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

and:

$\displaystyle |h(x)| \leq M$

then:

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

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:

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

[1-a]

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

[1-b]

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

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 $\displaystyle a$ and $\displaystyle b$, the follwoing is true:

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

[ii]

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

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

then the following inequality holds:

$\displaystyle a \cdot c < b \cdot d$

[iii]

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

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

and that:

$\displaystyle x_0 = x_1$

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

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

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 $\displaystyle \epsilon_1 > 0$ there is some $\displaystyle \delta_1 > 0$ such that:

[2]

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

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

[3]

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

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

[4-a]

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

[4-b]

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

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:

$\displaystyle \delta_1 = \delta_2$

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

$\displaystyle \epsilon_1 = (\epsilon_2 /M)$

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

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

[1-a]

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

[1-b]

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

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

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

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
• Aug 24th 2010, 03:49 PM
Plato
Can you show that for some $\displaystyle \delta>0$ you can make $\displaystyle \left| {g(x)} \right| < \frac{\varepsilon }{{M + 1}}$ if $\displaystyle |x|<\delta?$
• Aug 25th 2010, 01:55 PM
mfetch22
Quote:

Originally Posted by Plato
Can you show that for some $\displaystyle \delta>0$ you can make $\displaystyle \left| {g(x)} \right| < \frac{\varepsilon }{{M + 1}}$ if $\displaystyle |x|<\delta?$

Is this something that is true, and you are asking me to attempt to show so? Or are you asking me weather it is true or not? Assuming the first guess, then I'm not sure if I can show so; but I'll certainly give it a try, heres all my scratch work below (possibly including 'dead ends' ):

Quote:

SRATCH WORK:

I'm firstly going to make the stipulation that the $\displaystyle \varepsilon$ in you problem will be called $\displaystyle \epsilon_1$, and the $\displaystyle \delta$ be called $\displaystyle \delta_1$. So, I know from the problem that:

$\displaystyle 0 < |x| < \delta_1 \;\; \Rightarrow \;\; |g(x)| < \epsilon_2$

Soooo..... maybe... if I let:

$\displaystyle \epsilon_2 = \frac{\epsilon_1}{M + 1}$

Does that show what you were looking for?

Sorry, I'm new to proofs. I'm not exactly sure if I've shown, or even come remotely close to, what it is that you are asking to show.
• Aug 25th 2010, 01:56 PM
mfetch22
Quote:

Originally Posted by Plato
Can you show that for some $\displaystyle \delta>0$ you can make $\displaystyle \left| {g(x)} \right| < \frac{\varepsilon }{{M + 1}}$ if $\displaystyle |x|<\delta?$

I'm not sure how to show this, can you explain?