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dgamma3

  • one year ago

limit question. lim (x - > 5) 1/(x-5)^4

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  1. dgamma3
    • one year ago
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    \[\lim_{x \rightarrow 5} 1/(1+5)^4\]

  2. dgamma3
    • one year ago
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    woops. the 1 is an x

  3. kirbykirby
    • one year ago
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    It would be infinity

  4. kirbykirby
    • one year ago
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    oh

  5. dgamma3
    • one year ago
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    why so?

  6. kirbykirby
    • one year ago
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    Do you know L'Hopital's rule

  7. dgamma3
    • one year ago
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    nope. is it the only way?

  8. dgamma3
    • one year ago
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    the answer is infinity. just have no clue how to work it out.

  9. kirbykirby
    • one year ago
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    Ok what you can do is:

  10. kirbykirby
    • one year ago
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    wait i just noticed

  11. kirbykirby
    • one year ago
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    u mean it's x-5 in the denominator right? otherwise it wouldn't be infinity

  12. dgamma3
    • one year ago
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    yes, x-5

  13. kirbykirby
    • one year ago
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    \[\lim_{x \rightarrow 5} \frac{x}{(x-5)^4}\]\[=\lim_{x \rightarrow 5} \frac{x}{(x-5)^4}\]\[=\lim_{x \rightarrow 5} \frac{x}{(x-5)^2(x-5)^2}\]\[=\lim_{x \rightarrow 5} \frac{x}{(x-5)(x-5)(x-5)(x-5)}\]\[=\lim_{x \rightarrow 5} \frac{x/x}{(x/x-5/x)(x/x-5/x)(x/x-5/x)(x/x-5/x)}\]\[=\lim_{x \rightarrow 5} \frac{1}{(1-5/x)(1-5/x)(1-5/x)(1-5)}\]

  14. kirbykirby
    • one year ago
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    oops the last factor should also be (1-5/x)

  15. dgamma3
    • one year ago
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    awesome, thanks for your effort! just wondering line 6. why are you diving everything by x?

  16. kirbykirby
    • one year ago
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    I hope you see that the bottom factors are (1-1) when substituting x=5, so you get like "1/0" -> infinity

  17. kirbykirby
    • one year ago
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    hm it's true that was a useless step

  18. kirbykirby
    • one year ago
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    I was thinking that it was going to x->infinity at first, but i realized it was x->5 after lol

  19. dgamma3
    • one year ago
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    ahhh, yes. i see that. thank you. is that a legal move though? because you are changing the function?

  20. dgamma3
    • one year ago
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    for example, (x-5) doesn't equal (1-5/x)

  21. dgamma3
    • one year ago
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    sorry, i guess it is. just would have never thought of that!

  22. kirbykirby
    • one year ago
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    Nope :) If you divide the top and bottom by x, it is perfectly valid (it's like you are dividing by "1") Example: \[\frac{2}{3} = \frac{\frac{2}{x}}{\frac{3}{x}}=\frac{2}{x}*\frac{x}{3}=\frac{2}{3}\]

  23. kirbykirby
    • one year ago
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    I divided the numerator by x as well: I did x/x = 1 in the numerator, and divided the whole denominator by x

  24. kirbykirby
    • one year ago
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    if you divide the numerator, you MUST also divide the denominator!

  25. dgamma3
    • one year ago
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    awesome, this is just amazing. never would have figured out this.

  26. kirbykirby
    • one year ago
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    you will learn a neat theorem called "L'hopital's rule" later which will make the computation even easier :)

  27. dgamma3
    • one year ago
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    k. thanks for your time. im now 'a fan' haha.

  28. kirbykirby
    • one year ago
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    Hehe awesome ;) Good luck with everything

  29. dgamma3
    • one year ago
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    cheers

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