anonymous
  • anonymous
Prove limit using Epsilon Delta Defn lim [(x-1)(x+3)]/(x-2) =0 as x->1
Calculus1
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
katieb
  • katieb
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
I know \[\left| f(x)-L \right|<\epsilon \] and \[\left| x-1 \right| <\]
freckles
  • 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
zzr0ck3r
  • zzr0ck3r
Do you mean choose delta?

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

amistre64
  • 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
anonymous
  • 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
zzr0ck3r
  • 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. \(|x-1|<\delta=\frac{1}{2}\implies -1.50\) 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.

Looking for something else?

Not the answer you are looking for? Search for more explanations.