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anonymous
 one year ago
Prove limit using Epsilon Delta Defn
lim [(x1)(x+3)]/(x2) =0 as x>1
anonymous
 one year ago
Prove limit using Epsilon Delta Defn lim [(x1)(x+3)]/(x2) =0 as x>1

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I know \[\left f(x)L \right<\epsilon \] and \[\left x1 \right <\]

freckles
 one year ago
Best ResponseYou've already chosen the best response.0\[\frac{(x1)(x+3)}{x2} <\epsilon \\ x1 < \frac{\epsilonx2 }{x+3} \\ \text{ we can't express } \delta \text{ as a function of } x \\ \\ \text{ so let's look at } x1<1 \text{ which means } \\ 1<x1<1 \\ 0<x<2 \\ \text{ and so adding 3 on both sides } \\ 3<x+3<5 \\ \text{ or if instead we subtracted 2 on both sides } \\ 2<x2<0 \\ \text{ choose } x \text{ so that } \frac{\epsilon x2}{x+3} \text{ is at it's min value}\]

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.1Do you mean choose delta?

amistre64
 one year ago
Best ResponseYou've already chosen the best response.0we never spent much time on it in my classes .... but this seems like a good reference from my perspective. http://www.milefoot.com/math/calculus/limits/DeltaEpsilonProofs03.htm

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0This suggestion has a lot in common with what freckles wrote. First let \(t=x1\), so that as \(x\to1\) we have \(t\to0\). This doesn't change the problem (much); we still want to prove the same limit. \[\lim_{x\to1}\frac{(x1)(x+3)}{x2}=\lim_{t\to0}\frac{t(t+4)}{t1}=0\] According to the definition of the limit, we want to show that, given any \(\epsilon>0\), we can determine an appropriate \(\delta\) so that if \(t0=t<\delta\), then we must have \(\left\frac{t(t+4)}{t1}0\right=\left\frac{t(t+4)}{t1}\right<\epsilon\). Let's try to work backwards. The strategy behind \(\epsilon\delta\) proofs is to extract some inequality of the form \(t<(\text{some expression containing }\epsilon)\) from the desired \(f(t)L<\epsilon\). One thing we could do is to compare \(\frac{t(t+4)}{t1}\) to another rational expression such that the denominator can be canceled  namely, compare it to \(\frac{t(t1)}{t1}=t\). From the following plot, you can ascertain (at least for some interval containing \(t=0\)) that \[(*)\quad\quadt=\left\frac{t(t1)}{t1}\right\le\left\frac{t(t+4)}{t1}\right\] http://www.wolframalpha.com/input/?i=plot+Abs%5Bt%28t%2B4%29%5D%2C+Abs%5Bt%28t1%29%5D+over+2%2C2 You have to be careful about this statement, though, as it's not true if \(t<\frac{3}{2}\). To avoid this problem, we agree to fix \(t<1\), i.e. \(1<t<1\), where \((*)\) does hold. So, it follows that \[t=\left\frac{t(t1)}{t1}\right\le\left\frac{t(t+4)}{t1}\right<\epsilon\] This means that we can use \(\delta=\epsilon\), right? Wrong. Remember that \((*)\) is only true for certain \(t\), and of particular interest to us, it's true for \(t<1\). If \(\epsilon\) was large enough to force \(t\) outside of the interval \((1,1)\), we should like to fix \(\delta\) so that it defaults to a value we know works. That value is \(1\), since we know if \(t<1\), then \((*)\) certainly holds. All this to say that we should actually be using the smaller of the two, i.e. we choose \(\delta=\min\{1,\epsilon\}\). Finding \(\delta\) is the hard part, in my opinion, so I'll leave the rest to you. (Disclaimer: I'm writing this while pretty tired, if anyone sees some silly mistake in the reasoning here, please let me know.)

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.1So what I always try and do is get rid of the denominator is some way \(\dfrac{x1x+3}{x2}< \dfrac{1}{a}x1x+3\) would be sweet because then we can easily deal with the \(x+3\) part. I always start with \(\delta=\dfrac{1}{2}\) and then just play around. \(x1<\delta=\frac{1}{2}\implies 1.5<x1<0.5\implies 0.5<x2<1.5\) Now we can say \(f(x)0=\dfrac{x1x+3}{x2}<\dfrac{1}{\frac{1}{2}}x1x+3=2x1x+3\) when \(x1<\delta\). Now \(x1<1\implies 1<x1<1\implies 3<x+3<4\implies 3<x+3<4\) So now we have \[\dfrac{x1x+3}{x2}<\dfrac{1}{\frac{1}{2}}x1x+3=2x1x+3<2x1*4=8x1\] And we want all of this less than \(\epsilon\) , so now the proof. Let \(\epsilon >0\) be given. Set \(\delta = \min\{\frac{1}{2}, \frac{\epsilon}{8}\}\). Then \[\dfrac{x1x+3}{x2}<8x1<8\delta\le \epsilon \] Just thought I would throw in a different method.
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