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frx

  • 2 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

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  1. myininaya
    • 2 years ago
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    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
    • 2 years ago
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    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
    • 2 years ago
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    Ok. I couldn't read that little number.

  4. myininaya
    • 2 years ago
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    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
    • 2 years ago
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    E is epsilon d is delta by the way :)

  6. myininaya
    • 2 years ago
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    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
    • 2 years ago
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    Got it! Thanks a lot, really appreciate it! :D

  8. myininaya
    • 2 years ago
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    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
    • 2 years ago
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    Oh and wow. I can read that 3 on this computer but not that other one.

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