# HELP! Another limit question!

1. Oct 20, 2004

### johnnyICON

Hi, I have this question that I am not all to sure how to do. Actually, I don't know where to even start.

$\large r(x)=\frac{3x^2 - Sx - S^2}{2x^2 + Sx + S ^ 2}$ Where S represents some abitrary number, let's say 9 for an example.

I found the limit of the function to be $\frac{3x}{2}$

Then the question asks me to find a value of N that satisfies $|r(x) - \frac{3x}{2}| < 0.001$

I have no clue of where to start... I have these examples but they didn't help much.

2. Oct 21, 2004

### Galileo

So you're taking the limit of r(x) as x goes to ???

Furthermore, the limit (if it exists) is a number. Your limit has an x-dependence.

I assume you take the limit as x goes to infinity, since that would resemble your answer closely.
$$\lim_{x \rightarrow \infty}r(x)=\frac{3}{2}$$

Then I guess you must have $|r(x)-\frac{3}{2}|<0.001$ whenever $x>N$ for some number N.
I think N will depends on S. If S=9, you can check that N=1.000.000 for example easily does the job.

3. Oct 21, 2004

### arildno

Write:
$$|r(x)-\frac{3}{2}|=|\frac{3x^{2}-Sx-S^{2}}{2x^{2}+Sx+s^{2}}-3/2\frac{x^{2}+Sx+S^{2}}{x^{2}+Sx+S^{2}}|=$$
$$\frac{2}{x^{2}}|\frac{\frac{S}{x}+(\frac{S}{x})^{2}}{1+\frac{1}{2}(\frac{S}{x}+(\frac{S}{x})^{2})}|$$
1. Assume that x is so big that:
$$|\frac{S}{x}+(\frac{S}{x})^{2}|<1$$
Hence, we have:
$$|r(x)-\frac{3}{2}|\leq\frac{2}{x^{2}}\frac{1}{1-\frac{1}{2}}=(\frac{2}{x})^{2}$$
2. From this it is simple to find a value of x which gives you your estimate.
3. You are not quite finished, though:
Evidently, you must find an N so that BOTH your inequalities,
$$|\frac{S}{x}+(\frac{S}{x})^{2}|<1$$, $$(\frac{2}{x})^{2}<\epsilon$$
are fulfilled, i.e, choose your N to be the maximal value of the "separate N's" you derive for each inequality.
Your final N is therefore dependent on both $$S,\epsilon$$
($$\epsilon$$ is the stated error margin)

Last edited: Oct 21, 2004
4. Oct 21, 2004

### johnnyICON

Thanks, I don't have time to look at this yet. I'm late for class. But the limit was as $x$ approaches $infinity$. Sorry I should of mentioend that

5. Oct 21, 2004

### arildno

Sure, I know that the limit goes to infinity; the derived inequalities holds for "the absolute value of x greater than some number"

Last edited: Oct 21, 2004
6. Oct 21, 2004

### johnnyICON

Sorry, the message was for the post prior to yours... the one by... Galileo. Thanks for you help! I still don't have time to look at this... I'll give it a gander when I get home.