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anonymous

  • one year ago

I NEED HELP HERE

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  1. anonymous
    • one year ago
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    @Michele_Laino PLEASE POST IT HERE WHEN YOU ARE THROUGH

  2. anonymous
    • one year ago
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  3. anonymous
    • one year ago
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    CAN ANYONE HELP PLEASE

  4. Jack1
    • one year ago
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    cant open the file... u have a screenshot dude?

  5. anonymous
    • one year ago
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    no

  6. anonymous
    • one year ago
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    http://prntscr.com/83uvhy

  7. anonymous
    • one year ago
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    Screenie for @jack1

  8. anonymous
    • one year ago
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    @Michele_Laino

  9. anonymous
    • one year ago
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    http://prntscr.com/83uvhy

  10. anonymous
    • one year ago
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    @Michele_Laino please help with this open question

  11. Michele_Laino
    • one year ago
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    by definition of the limit value, we can write this: \[\Large \forall \varepsilon ,\exists \delta |{\left\| {x - {x_0}} \right\|_{{\mathbb{R}^n}}} \leqslant \delta \Rightarrow \left| {f\left( x \right) - L} \right| \leqslant \varepsilon \]

  12. Michele_Laino
    • one year ago
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    that is the definition of the limit value of f(x) when x--->x_0

  13. Michele_Laino
    • one year ago
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    similarly for g(x): \[\Large \forall \varepsilon ,\exists \delta \;|\;\;{\left\| {x - {x_0}} \right\|_{{\mathbb{R}^n}}} \leqslant \delta \Rightarrow \left| {g\left( x \right) - M} \right| \leqslant \varepsilon \]

  14. Michele_Laino
    • one year ago
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    where the symbol: \[\Large {\left\| x -x_0 \right\|_{{\mathbb{R}^n}}}\] stands for the norm of R^n

  15. Michele_Laino
    • one year ago
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    of course I mean \[\Large \forall \varepsilon > 0,\exists \delta > 0\]

  16. Michele_Laino
    • one year ago
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    namely, \epsilon and \delta are both positive numbers

  17. Michele_Laino
    • one year ago
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    tell me when I may continue

  18. anonymous
    • one year ago
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    ok continue sir

  19. Michele_Laino
    • one year ago
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    now, we can show, that if f(x)--->L, when x--->x_0, then: (af)(x)---> aL, when x--->x_0, here is the proof:

  20. Michele_Laino
    • one year ago
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    by definition, of limit value, we can write: \[\Large \forall \varepsilon > 0,\exists \delta > 0|{\left\| {x - {x_0}} \right\|_{{\mathbb{R}^n}}} \leqslant \delta \Rightarrow \left| {f\left( x \right) - L} \right| \leqslant \varepsilon /\left| a \right|\]

  21. Michele_Laino
    • one year ago
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    and therefore, we have: \[\large \left| {\left( {af} \right)\left( x \right) - aL} \right| = \left| {af\left( x \right) - aL} \right| = \left| a \right|\left| {f\left( x \right) - L} \right| \leqslant \left| a \right| \cdot \left( {\varepsilon /\left| a \right|} \right) = \varepsilon \]

  22. Michele_Laino
    • one year ago
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    \[\Large \begin{gathered} \left| {\left( {af} \right)\left( x \right) - aL} \right| = \left| {af\left( x \right) - aL} \right| = \hfill \\ \hfill \\ = \left| a \right|\left| {f\left( x \right) - L} \right| \leqslant \left| a \right| \cdot \left( {\varepsilon /\left| a \right|} \right) = \varepsilon \hfill \\ \end{gathered} \]

  23. Michele_Laino
    • one year ago
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    so we are done!

  24. Michele_Laino
    • one year ago
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    similarly, if g(x)--->M, when x--->x_0, then, we have: (bg)(x)---> bM, when x--->x_0

  25. anonymous
    • one year ago
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    but we have addition, subtraction, multiplication and division?

  26. Michele_Laino
    • one year ago
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    yes! I know, at the moment I have proven that, if: f--->L then af--->aL

  27. Michele_Laino
    • one year ago
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    next I will prove that if f--->L and g--->M, then: f+g--->L+M

  28. anonymous
    • one year ago
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    ok sir . so is that for 3I in the question?

  29. Michele_Laino
    • one year ago
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    yes! exactly

  30. Michele_Laino
    • one year ago
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    here is the proof:

  31. Michele_Laino
    • one year ago
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    by hypothesis, we can write this: \[\Large \begin{gathered} \forall \varepsilon > 0,\exists \delta > 0|{\left\| {x - {x_0}} \right\|_{{\mathbb{R}^n}}} \leqslant \delta \Rightarrow \left| {f\left( x \right) - L} \right| \leqslant \varepsilon /2 \hfill \\ \hfill \\ \forall \varepsilon > 0,\exists \delta > 0\;|\;\;{\left\| {x - {x_0}} \right\|_{{\mathbb{R}^n}}} \leqslant \delta \Rightarrow \left| {g\left( x \right) - M} \right| \leqslant \varepsilon /2 \hfill \\ \end{gathered} \]

  32. Michele_Laino
    • one year ago
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    then we have: \[\Large \begin{gathered} \left| {f\left( x \right) + g\left( x \right) - \left( {L + M} \right)} \right| = \left| {f\left( x \right) - L + g\left( x \right) - M} \right| \leqslant \hfill \\ \hfill \\ \leqslant \left| {f\left( x \right) - L} \right| + \left| {g\left( x \right) - M} \right| \leqslant \frac{\varepsilon }{2} + \frac{\varepsilon }{2} = \varepsilon \hfill \\ \end{gathered} \]

  33. Michele_Laino
    • one year ago
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    so, we are done!

  34. Michele_Laino
    • one year ago
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    now, the property i) is completely proven

  35. anonymous
    • one year ago
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    ok sir

  36. Michele_Laino
    • one year ago
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    next I give the proof of property ii)

  37. Michele_Laino
    • one year ago
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    here, by hypothesis, we can write this: \[\large \begin{gathered} \forall \varepsilon > 0,\exists \delta > 0|{\left\| {x - {x_0}} \right\|_{{\mathbb{R}^n}}} \leqslant \delta \Rightarrow \left| {f\left( x \right) - L} \right| \leqslant \varepsilon /2K \hfill \\ \hfill \\ \forall \varepsilon > 0,\exists \delta > 0\;|\;\;{\left\| {x - {x_0}} \right\|_{{\mathbb{R}^n}}} \leqslant \delta \Rightarrow \left| {g\left( x \right) - M} \right| \leqslant \varepsilon /2K \hfill \\ \hfill \\ K = \max \left( {1 + \left| M \right|,1 + \left| L \right|} \right) \hfill \\ \end{gathered} \]

  38. Michele_Laino
    • one year ago
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    herefore, we have the subsequent steps: \[\large \begin{gathered} \left| {f\left( x \right)g\left( x \right) - LM} \right| = \left| {f\left( x \right)g\left( x \right) - Lg\left( x \right) + Lg\left( x \right) - LM} \right| = \hfill \\ \hfill \\ = \left| {\left( {f\left( x \right) - L} \right)g\left( x \right) + L\left( {g\left( x \right) - M} \right)} \right| \leqslant \hfill \\ \hfill \\ \leqslant \left| {g\left( x \right)} \right| \cdot \left| {f\left( x \right) - L} \right| + \left| L \right| \cdot \left| {g\left( x \right) - M} \right| \hfill \\ \end{gathered} \]

  39. Michele_Laino
    • one year ago
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    I have added and subtracted the quantity L*g(x)

  40. Michele_Laino
    • one year ago
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    next we note that: \[\left| {g\left( x \right)} \right| = \left| {g\left( x \right) - M + M} \right| \leqslant \left| {g\left( x \right) - M} \right| + \left| M \right| \leqslant 1 + \left| M \right|\]

  41. Michele_Laino
    • one year ago
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    \[\large \left| {g\left( x \right)} \right| = \left| {g\left( x \right) - M + M} \right| \leqslant \left| {g\left( x \right) - M} \right| + \left| M \right| \leqslant 1 + \left| M \right|\]

  42. Michele_Laino
    • one year ago
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    more precisely: \[\Large \begin{gathered} \left| {g\left( x \right)} \right| = \left| {g\left( x \right) - M + M} \right| \leqslant \left| {g\left( x \right) - M} \right| + \left| M \right| \leqslant \hfill \\ \hfill \\ \leqslant \varepsilon + \left| M \right| \leqslant 1 + \left| M \right| \hfill \\ \end{gathered} \]

  43. Michele_Laino
    • one year ago
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    so, we can write: \[\large \begin{gathered} \left| {f\left( x \right)g\left( x \right) - LM} \right| = \left| {f\left( x \right)g\left( x \right) - Lg\left( x \right) + Lg\left( x \right) - LM} \right| = \hfill \\ \hfill \\ = \left| {\left( {f\left( x \right) - L} \right)g\left( x \right) + L\left( {g\left( x \right) - M} \right)} \right| \leqslant \hfill \\ \hfill \\ \leqslant \left| {g\left( x \right)} \right| \cdot \left| {f\left( x \right) - L} \right| + \left| L \right| \cdot \left| {g\left( x \right) - M} \right| \leqslant \hfill \\ \hfill \\ \leqslant \left( {1 + \left| M \right|} \right) \cdot \left| {f\left( x \right) - L} \right| + \left| L \right| \cdot \left| {g\left( x \right) - M} \right| \leqslant \hfill \\ \hfill \\ \leqslant K\left( {\left| {f\left( x \right) - L} \right| + \left| {g\left( x \right) - M} \right|} \right) \leqslant \hfill \\ \hfill \\ \leqslant K\left( {\frac{\varepsilon }{{2K}} + \frac{\varepsilon }{{2K}}} \right) = \varepsilon \hfill \\ \end{gathered} \]

  44. Michele_Laino
    • one year ago
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    so, we are done

  45. anonymous
    • one year ago
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    thanks. great. so we are left with one now

  46. Michele_Laino
    • one year ago
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    yes! we have to prove the property iii)

  47. Michele_Laino
    • one year ago
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    again, by hypothesis, we can write this: \[\large \begin{gathered} \forall \varepsilon > 0,\exists \delta > 0|{\left\| {x - {x_0}} \right\|_{{\mathbb{R}^n}}} \leqslant \delta \Rightarrow \left| {f\left( x \right) - L} \right| \leqslant \varepsilon /2K \hfill \\ \hfill \\ \forall \varepsilon > 0,\exists \delta > 0\;|\;\;{\left\| {x - {x_0}} \right\|_{{\mathbb{R}^n}}} \leqslant \delta \Rightarrow \left| {g\left( x \right) - M} \right| \leqslant \varepsilon /2K \hfill \\ \hfill \\ K = \max \left( {\frac{{\left| M \right|}}{{\left| {M - \varepsilon } \right|\left| M \right|}},\frac{{\left| L \right|}}{{\left| {M - \varepsilon } \right|\left| M \right|}}} \right) \hfill \\ \end{gathered} \]

  48. Michele_Laino
    • one year ago
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    then we have: \[\large \begin{gathered} \left| {\frac{{f\left( x \right)}}{{g\left( x \right)}} - \frac{L}{M}} \right| = \left| {\frac{{f\left( x \right)M - g\left( x \right)L}}{{g\left( x \right)M}}} \right| = \hfill \\ \hfill \\ = \left| {\frac{{f\left( x \right)M - LM + LM - g\left( x \right)L}}{{g\left( x \right)M}}} \right| = \hfill \\ \hfill \\ = \frac{{M\left( {f\left( x \right) - L} \right) + L\left( {M - g\left( x \right)} \right)}}{{g\left( x \right)M}} \leqslant \hfill \\ \hfill \\ \leqslant \frac{{\left| M \right|\left| {f\left( x \right) - L} \right| + \left| L \right|\left| {M - g\left( x \right)} \right|}}{{\left| {g\left( x \right)} \right|\left| M \right|}} \hfill \\ \end{gathered} \]

  49. Michele_Laino
    • one year ago
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    I have added and subtracted the quantity L*M

  50. Michele_Laino
    • one year ago
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    now, by hypothesis, we can write: \[\Large g\left( x \right) \geqslant M - \frac{\varepsilon }{{2K}} \geqslant M - 1\] or: \[\Large \frac{1}{{g\left( x \right)}} \leqslant \frac{1}{{M - 1}} \Rightarrow \frac{1}{{\left| {g\left( x \right)} \right|}} \leqslant \frac{1}{{\left| {M - 1} \right|}}\]

  51. Michele_Laino
    • one year ago
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    therefore, we have: \[\large \begin{gathered} \left| {\frac{{f\left( x \right)}}{{g\left( x \right)}} - \frac{L}{M}} \right| = \left| {\frac{{f\left( x \right)M - g\left( x \right)L}}{{g\left( x \right)M}}} \right| = \hfill \\ \hfill \\ = \left| {\frac{{f\left( x \right)M - LM + LM - g\left( x \right)L}}{{g\left( x \right)M}}} \right| = \hfill \\ \hfill \\ = \frac{{M\left( {f\left( x \right) - L} \right) + L\left( {M - g\left( x \right)} \right)}}{{g\left( x \right)M}} \leqslant \hfill \\ \hfill \\ \leqslant \frac{{\left| M \right|\left| {f\left( x \right) - L} \right| + \left| L \right|\left| {M - g\left( x \right)} \right|}}{{\left| {g\left( x \right)} \right|\left| M \right|}} \leqslant \hfill \\ \hfill \\ \leqslant \frac{{\left| M \right|\left| {f\left( x \right) - L} \right| + \left| L \right|\left| {M - g\left( x \right)} \right|}}{{\left| {M - 1} \right|\left| M \right|}} \leqslant \hfill \\ \hfill \\ \leqslant K\left( {\left| {f\left( x \right) - L} \right| + \left| {M - g\left( x \right)} \right|} \right) \leqslant \hfill \\ \hfill \\ \leqslant K\left( {\frac{\varepsilon }{{2K}} + \frac{\varepsilon }{{2K}}} \right) = \varepsilon \hfill \\ \end{gathered} \]

  52. Michele_Laino
    • one year ago
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    oops.. the right definition of K is: \[\Large K = \max \left( {\frac{{\left| M \right|}}{{\left| {M - 1} \right|\left| M \right|}},\frac{{\left| L \right|}}{{\left| {M - 1} \right|\left| M \right|}}} \right)\]

  53. Michele_Laino
    • one year ago
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    so we are done

  54. anonymous
    • one year ago
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    ok, thanks . will study it please do not delete

  55. anonymous
    • one year ago
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    please post number 2b solution steps here sir. thanks @Michele_Laino

  56. Michele_Laino
    • one year ago
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    please, attach that exercise, using the "Attach File" button

  57. Michele_Laino
    • one year ago
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    more precisely in the last proof, we can suppose, more simply: \[\Large K = \frac{1}{{\left| {M - 1} \right|}}\]

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