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So I know how to find the $\delta$ in $f(x)$ that can be factored into a degree of $1$, or I can solve it by solving $L + \epsilon = f(x) $ then finding the distance from $a$. But how do I find the delta analytically for an $f(x)$ for example a degree $2$.

Example: $f(x) = x^2$; $a =3$; $L = 9$; $\epsilon = 0.5$ $$ 0 < |x+3|<\delta $$ $$ |x^2-9| < 0.5 $$ $$ |x+3| |x-3| < 0.5$$ I get stuck at this part when solving the epsilon side.

$$ |x+3| |x-3| < 0.5$$ I can't factor $|x+3|$ like in functions of degree $1$. Can someone please explain?

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I was assuming you were taking the limit as $x \to 3$. This should be made explicit. It is required for the replacement of $|x+3|$ by $4$ once you choose a maximum $\delta$ of $1$ –  Ross Millikan Jan 24 '13 at 20:13

2 Answers 2

up vote 2 down vote accepted

If $0\le \delta \le 1$, then $|x-3|\lt \delta$, then $x+3\lt 7$. Thus if furthermore $|x-3|\lt \frac{\epsilon}{7}$, then $|x^2-9|\lt \epsilon$.

So if we pick $\delta=\min(1,\epsilon/7)$, then $|x^2-9|\lt \epsilon$ whenever $|x-3|\lt \delta$.

Here we did not find the cheapest $\delta$ that works. Proving that the limit exists does not require that we find the cheapest $\delta$, given $\epsilon$. It only requires that we show there is a $\delta$.

Remark: If you really want to find the cheapest $\delta$, for say $\epsilon=0.5$, you need to solve the inequality $|x^2-9|\lt 0.5$. Separate this inequality into two cases (i) |x|\ge 3$ and (ii) |x|\lt 3$. for (i), we are looking at the inequality $x^2\lt 9.5$, and for (ii) we are looking at $x^2\gt 8.5$. Now use the calculator.

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@RossMillikan: Annoying to find I am addition-challenged. Thought the debility was limited to subtraction. –  André Nicolas Jan 25 '13 at 0:19

You can bound $|x+3|$ by taking some preliminary $\delta$ of convenience. For example, just say you will start with $\delta \le 1$. Then $|x+3| \le 7$. Since you just have to prove your error is less than $\epsilon$, you can say $|x+3||x-3| \le 7 |x-3|$ and continue. Then report $\delta$ as the smaller of the result of your calculation (which it will be here) or $1$.

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On what basis can I consider the preliminary $\delta$ to be convenient? Any tips on how to choose one? –  Men Jan 24 '13 at 18:00
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@vincentbelkin: It really doesn't matter. $1$ is convenient. You are just trying to bound the factors that don't go to zero. In this case, if you used $1$ you would find $\delta \le \frac {\epsilon}4$, while if you used $\frac 12$ you would get $\delta \le \frac {\epsilon}{3.5}$, but you just are trying to show that some $\delta$ works. –  Ross Millikan Jan 24 '13 at 18:09
    
I'm trying to solve a similar problem but can't seem to figure out why $|x+3| \le 4$ if I choose $\delta \le 1$? –  Charles Khunt Jan 24 '13 at 20:06
    
Wow I must be really dumb as I still can't wrap my head around this.. so basically what you're saying is that $|x-3| \lt 1 \Rightarrow |x+3| \lt 4$? –  Charles Khunt Jan 24 '13 at 20:45
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I must be missing something fundamental here.. What I did is: $|x-3| \lt 1 \Rightarrow 2 \lt x \lt 4 \Rightarrow 2 + 3 \lt x + 3 \lt 4 + 3 \Rightarrow 5 \lt x + 3 \lt 7$.. I don't know if that even makes sense. Do you mind showing some algebraic steps? –  Charles Khunt Jan 24 '13 at 23:44

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