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frx
Group Title
Let \[f(x) = x^2  2x +1 \] use the epsilondelta definition to show that: \[\lim_{x \rightarrow 3} f(x) = 4\]
I've come up with this:
\[( x1)^2  4 < \epsilon \]
\[ x3 < \sqrt{\epsilon } \]
Set delta as the squareroot of epsilon:
\[ x3 < \delta =\sqrt{\epsilon } \]
How do I finish this, I get it all wrong by taking the left expresson to the second power
 one year ago
 one year ago
frx Group Title
Let \[f(x) = x^2  2x +1 \] use the epsilondelta definition to show that: \[\lim_{x \rightarrow 3} f(x) = 4\] I've come up with this: \[( x1)^2  4 < \epsilon \] \[ x3 < \sqrt{\epsilon } \] Set delta as the squareroot of epsilon: \[ x3 < \delta =\sqrt{\epsilon } \] How do I finish this, I get it all wrong by taking the left expresson to the second power
 one year ago
 one year ago

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myininaya Group TitleBest ResponseYou've already chosen the best response.1
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>2^22(2)+1=44+1=0+1=1
 one year ago

frx Group TitleBest ResponseYou've already chosen the best response.0
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 \[ x \rightarrow 3\]
 one year ago

myininaya Group TitleBest ResponseYou've already chosen the best response.1
Ok. I couldn't read that little number.
 one year ago

myininaya Group TitleBest ResponseYou've already chosen the best response.1
Thanks. So you have x^22x+14<E x^22x3<E (x3)(x+1)<E (x3)(x+1)<E x3<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
 one year ago

myininaya Group TitleBest ResponseYou've already chosen the best response.1
E is epsilon d is delta by the way :)
 one year ago

myininaya Group TitleBest ResponseYou've already chosen the best response.1
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 :(
 one year ago

frx Group TitleBest ResponseYou've already chosen the best response.0
Got it! Thanks a lot, really appreciate it! :D
 one year ago

myininaya Group TitleBest ResponseYou've already chosen the best response.1
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}
 one year ago

myininaya Group TitleBest ResponseYou've already chosen the best response.1
Oh and wow. I can read that 3 on this computer but not that other one.
 one year ago
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