Question

Prove limit using Epsilon Delta Defn

lim [(x-1)(x+3)]/(x-2) =0 as x->1

Answer 1

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

Answer 2

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

Answer 3

Do you mean choose delta?

Answer 4

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

Answer 5

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.)

Answer 6

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.