Proving a limit of at infinity based on certain given stipulations?

**mfetch22** · August 20th 2010, 10:42 AM

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:

LIMIT PROBLEM

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

and

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

then prove that:

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

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

MY ATTEMPT AT PROVING THE LIMIT

By the definition of a limit, we know that:

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

Means that for every:
$\epsilon_2 > 0$ ,theres is a $\delta_1 > 0$ such that:

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

implies:

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

The above inequality implies that:

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

Also, since:

$0 < \epsilon_1 < f(x)$
Then it follows that (by multiplying $\epsilon_1$ to both sides of the g(x) inequality):

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

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

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

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

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

Implies that, for any $N$ we have:

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

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 :

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

So letting:

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

and

$\delta_1 = \delta_2$

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

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

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

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

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

Remeber that:

$0 < \epsilon_1 < f(x)$

So:

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

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

Which shows that:

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

For:

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

and

$\delta_1 = \delta_2$

FINNALY, this all means that:

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

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

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 $0 \leq |a|$ for all $a$ ) Thanks in advance

**Opalg** · August 20th 2010, 11:14 AM

That argument is essentially correct. The only thing I would criticise is that you need to be clear about what is given and what is subsequently chosen. The quantity $\epsilon_1$ is given in the statement of the problem. The quantity N is then given as a "target" in the definition of the limit being infinity. The quantity that is then chosen by you during the course of the proof is $\epsilon_2$ . So in the middle of the proof it would be better to write "Let $\epsilon_2 = \epsilon_1/N$ " rather than letting $N = \epsilon_1/\epsilon_2$ .

**mfetch22** · August 20th 2010, 05:11 PM

Originally Posted by Opalg

That argument is essentially correct. The only thing I would criticise is that you need to be clear about what is given and what is subsequently chosen. The quantity $\epsilon_1$ is given in the statement of the problem. The quantity N is then given as a "target" in the definition of the limit being infinity. The quantity that is then chosen by you during the course of the proof is $\epsilon_2$ . So in the middle of the proof it would be better to write "Let $\epsilon_2 = \epsilon_1/N$ " rather than letting $N = \epsilon_1/\epsilon_2$ .

Thank you for that clarification. But I'm curious about one detail. I trust your judgement extremely more then mine concerning these matters of mathematical proofs, since I'm quite new to learning real analysis; but I just wanted to make 100% sure of my understanding.

You implied that it'd be more appropriate "to write 'Let $\epsilon_2 = \epsilon_1 / N \;$ rather then $N = \epsilon_1 / \epsilon_2$ ". Remebering what was said in the textbook I've been using over the summer, I thought that (and this may just be my mistunderstanding of the limit chapter) but I thought that the definition of a limit, and the given information implies that:

$\epsilon_2 = \epsilon (\delta_1)$

Where $\; \epsilon(\delta_1) \;$ is a function of $\delta_1$

?

What I mean is, since the problem provided that:

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

Or in other words:

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

And since we are given the fact that the left side of the statement above implies the right side, is it neccesarily valid to assume from this (assuming from the given information that is) that $\epsilon_2$ can be written as a function of $\delta_1$ ?

If this is a valid assumption, then would it be correct or incorrect to still just define:

$N = \epsilon_1 / \epsilon_2$

Since the infromation given in the problem implies that $\epsilon_2$ already has a definition?

Or, since a definition was not formmally given at any point in the proof (it was only implied to exist, if I was correct in my assumption), is it required by rigor to place within the proof a definition of $\epsilon_2$ ? Which, as you said would be:

$\epsilon_2 = \epsilon_1 / N$

?

Thank you for your help

**Opalg** · August 21st 2010, 12:18 AM

Originally Posted by mfetch22

I thought that the definition of a limit, and the given information implies that:

$\epsilon_2 = \epsilon (\delta_1)$

Where $\; \epsilon(\delta_1) \;$ is a function of $\delta_1$

?

Wrong way round! The definition of a limit says that given $\epsilon>0$ there exists $\delta>0$ such that (blah blah blah). So $\delta$ is a function of $\epsilon$ , not the other way round.

In your problem, you want to show that if x is sufficiently close to a, then f(x)/|g(x)| can be made larger than some given N. That will be the case provided that |g(x)| can be made less than $\epsilon_2 = \epsilon_1/N$ . And the definition of $\lim_{x\to a}g(x)=0$ says that this in turn can be achieved provided that $|x-a|<\delta_2$ for some $\delta_2$ depending on $\epsilon_2$ .

**mfetch22** · August 21st 2010, 08:12 AM

Originally Posted by Opalg

Wrong way round! The definition of a limit says that given $\epsilon>0$ there exists $\delta>0$ such that (blah blah blah). So $\delta$ is a function of $\epsilon$ , not the other way round.

In your problem, you want to show that if x is sufficiently close to a, then f(x)/|g(x)| can be made larger than some given N. That will be the case provided that |g(x)| can be made less than $\epsilon_2 = \epsilon_1/N$ . And the definition of $\lim_{x\to a}g(x)=0$ says that this in turn can be achieved provided that $|x-a|<\delta_2$ for some $\delta_2$ depending on $\epsilon_2$ .

Ohhhhhhhhhh! I seee! Your completely correct! For some reason it seems to make much more sense now, in the way you worded it above. Thank you for your assistance

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