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anonymous

  • one year ago

can someone tell me how to prove for infinite limit with epsilon and delta or N and M?

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  1. abb0t
    • one year ago
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    Yes, if you gave us an example, we could.

  2. abb0t
    • one year ago
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    What's the function?

  3. anonymous
    • one year ago
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    l\[\lim_{x \rightarrow \infty}3x^{2}-x-2/5x^{2}+4x+1\]

  4. ganeshie8
    • one year ago
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    \[\lim_{x \rightarrow \infty}~\dfrac{3x^{2}-x-2}{5x^{2}+4x+1}\] like that ?

  5. anonymous
    • one year ago
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    Yes, like that

  6. ganeshie8
    • one year ago
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    so what does it evaluate to ?

  7. ganeshie8
    • one year ago
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    you need to know the limit before starting to prove it using epsilon-delta

  8. anonymous
    • one year ago
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    i know the limit it's 3/5

  9. anonymous
    • one year ago
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    but the prove is quite confusing.

  10. ganeshie8
    • one year ago
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    its easy once you know how to play the game

  11. anonymous
    • one year ago
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    i guess... so :)

  12. ganeshie8
    • one year ago
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    Let \(f(x) =\dfrac{3x^{2}-x-2}{5x^{2}+4x+1} \) given any positive number \(\epsilon\), you have to show that there exists some \(x\) value, after which the difference between \(f(x)\) and \(\dfrac{3}{5}\) is less than \(\epsilon\).

  13. ganeshie8
    • one year ago
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    Given any positive number \(\epsilon\), prove that there exists some \(x\) value beyond which below is true : \(\left|\dfrac{3x^2-x-2}{5x^2+4x+1}-\dfrac{3}{5}\right|\lt \epsilon\)

  14. ganeshie8
    • one year ago
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    simplify and try expressing \(x\) in terms of \(\epsilon\)

  15. ganeshie8
    • one year ago
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    @kal1921

  16. anonymous
    • one year ago
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    simplify? sorry where is delta?

  17. zzr0ck3r
    • one year ago
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    Simplify so that you can find the bound on delta. You have to work backwards

  18. ganeshie8
    • one year ago
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    yeah what we're doing is just scratch work to express \(x\) in terms of \(\epsilon\)

  19. anonymous
    • one year ago
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    \[-\epsilon< equation<\epsilon\] i get this...what is next?

  20. ganeshie8
    • one year ago
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    for \(x\gt 1\), we have : \( \left|\dfrac{3x^2-x-2}{5x^2+4x+1}-\dfrac{3}{5}\right|\\~\\ =\left|\dfrac{17x+13}{5(5x^2+4x+1)}\right|\\~\\ \lt\left|\dfrac{17x+13}{25x^2}\right|\\~\\ \lt\left|\dfrac{17x+17x}{25x^2}\right|\\~\\ =\left|\dfrac{34}{25x}\right|\\~\\ \) \(\dfrac{34}{25x} \lt \epsilon \implies x \gt \dfrac{34}{25\epsilon}\) therefore \(x = \dfrac{34}{25\epsilon}\) works, now you can start the proof.

  21. anonymous
    • one year ago
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    sorry, how can i relate x and delta... if it x wasn't approaching to infinity i would say /x-a/>delta

  22. ganeshie8
    • one year ago
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    The definition for limits at infinity is slightly different : \(\lim\limits_{x\to\infty}f(x)=L\) means that for "every" \(\epsilon\gt 0\), there exists some \(x\) value, \(N\), such that \(|f(x)-L|\lt\epsilon\) for all \(x\gt N\).

  23. ganeshie8
    • one year ago
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    In present problem, we have cooked up that \(N\) to be \(\dfrac{34}{25\epsilon}\)

  24. ganeshie8
    • one year ago
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    you choose "any" positive \(\epsilon\), the value \(|f(x)-\frac{3}{5}|\) will always be less than that chosen \(\epsilon\) whenever \(x\) is greater than \(\dfrac{34}{25\epsilon}\)

  25. ganeshie8
    • one year ago
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    In other words, we can make \(f(x)\) as close to \(\frac{3}{5}\) as we want by making \(x\) sufficiently large.

  26. ganeshie8
    • one year ago
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    I think this is same as the N and M definition that you were refering to in main question

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