## anonymous one year ago Prove limit using Epsilon Delta Defn lim [(x-1)(x+3)]/(x-2) =0 as x->1

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1. anonymous

I know $\left| f(x)-L \right|<\epsilon$ and $\left| x-1 \right| <$

2. freckles

$|\frac{(x-1)(x+3)}{x-2}| <\epsilon \\ |x-1| < \frac{\epsilon|x-2| }{|x+3|} \\ \text{ we can't express } \delta \text{ as a function of } x \\ \\ \text{ so let's look at } |x-1|<1 \text{ which means } \\ -1<x-1<1 \\ 0<x<2 \\ \text{ and so adding 3 on both sides } \\ 3<x+3<5 \\ \text{ or if instead we subtracted 2 on both sides } \\ -2<x-2<0 \\ \text{ choose } x \text{ so that } \frac{\epsilon |x-2|}{|x+3|} \text{ is at it's min value}$

3. zzr0ck3r

Do you mean choose delta?

4. amistre64

we 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

5. anonymous

This suggestion has a lot in common with what freckles wrote. First let $$t=x-1$$, so that as $$x\to1$$ we have $$t\to0$$. This doesn't change the problem (much); we still want to prove the same limit. $\lim_{x\to1}\frac{(x-1)(x+3)}{x-2}=\lim_{t\to0}\frac{t(t+4)}{t-1}=0$ 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 $$|t-0|=|t|<\delta$$, then we must have $$\left|\frac{t(t+4)}{t-1}-0\right|=\left|\frac{t(t+4)}{t-1}\right|<\epsilon$$. 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 $$\frac{t(t+4)}{t-1}$$ to another rational expression such that the denominator can be canceled - namely, compare it to $$\frac{t(t-1)}{t-1}=t$$. From the following plot, you can ascertain (at least for some interval containing $$t=0$$) that $(*)\quad\quad|t|=\left|\frac{t(t-1)}{t-1}\right|\le\left|\frac{t(t+4)}{t-1}\right|$ http://www.wolframalpha.com/input/?i=plot+Abs%5Bt%28t%2B4%29%5D%2C+Abs%5Bt%28t-1%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 $|t|=\left|\frac{t(t-1)}{t-1}\right|\le\left|\frac{t(t+4)}{t-1}\right|<\epsilon$ 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.)

6. zzr0ck3r

So what I always try and do is get rid of the denominator is some way $$\dfrac{|x-1||x+3|}{|x-2|}< \dfrac{1}{a}|x-1||x+3|$$ 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. $$|x-1|<\delta=\frac{1}{2}\implies -1.5<x-1<-0.5\implies 0.5<|x-2|<1.5$$ Now we can say $$|f(x)-0|=\dfrac{|x-1||x+3|}{|x-2|}<\dfrac{1}{\frac{1}{2}}|x-1||x+3|=2|x-1||x+3|$$ when $$|x-1|<\delta$$. Now $$|x-1|<1\implies -1<x-1<1\implies 3<x+3<4\implies 3<|x+3|<4$$ So now we have $\dfrac{|x-1||x+3|}{|x-2|}<\dfrac{1}{\frac{1}{2}}|x-1||x+3|=2|x-1||x+3|<2|x-1|*4=8|x-1|$ 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 $\dfrac{|x-1||x+3|}{|x-2|}<8|x-1|<8\delta\le \epsilon$ Just thought I would throw in a different method.