Finding a Value of N for a Limit Question: How to Solve |r(x) - 3x/2| < 0.001

[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}| &lt; 0.001

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

 

[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}|&lt;0.001 whenever x&gt;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.

 

[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}|&lt;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}|&lt;1, (\frac{2}{x})^{2}&lt;\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)

 

[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

 

[arildno]

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

 

[johnnyICON]

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

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.