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|

Question

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

myininaya · Answer

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

anonymous · Answer

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 $$ x \rightarrow 3$$

myininaya · Answer

Ok. I couldn't read that little number.

myininaya · Answer

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|

myininaya · Answer

E is epsilon
d is delta 
by the way :)

myininaya · Answer

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 :(

anonymous · Answer

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

myininaya · Answer

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}

myininaya · Answer

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