## frx Group Title Let $f(x) = x^2 - 2x +1$ use the epsilon-delta definition to show that: $\lim_{x \rightarrow 3} f(x) = 4$ I've come up with this: $|( x-1)^2 - 4| < \epsilon$ $| x-3| < \sqrt{\epsilon }$ Set delta as the squareroot of epsilon: $| x-3| < \delta =\sqrt{\epsilon }$ How do I finish this, I get it all wrong by taking the left expresson to the second power 2 years ago 2 years ago

1. myininaya

How do you have the limit is 4? You won't be able to prove what you said. That does x->2 right? So f->2^2-2(2)+1=4-4+1=0+1=1

2. frx

Well, that the limit is 4 was given in the assignment and $f(x)=x2−2x+=4$. where x=3 I think you got the limit wrong it's $x \rightarrow 3$

3. myininaya

Ok. I couldn't read that little number.

4. myininaya

Thanks. So you have |x^2-2x+1-4|<E |x^2-2x-3|<E |(x-3)(x+1)|<E |(x-3)||(x+1)|<E |x-3|<E/|x+1| I would choose d<1 And then choose the number in the interval of x that gives me the smallest E/|x+1|

5. myininaya

E is epsilon d is delta by the way :)

6. myininaya

I hope that helps. I will try to check back with you later. I have to go for now, but I return in like 2 hours :(

7. frx

Got it! Thanks a lot, really appreciate it! :D

8. myininaya

Great remember when showing that limit above you have to say let d=min{1, E/|x+1| where you have chosen x based off letting d<1}

9. myininaya

Oh and wow. I can read that 3 on this computer but not that other one.