## frx 3 years ago 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

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.