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anonymous

  • one year ago

Just a small question...

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  1. anonymous
    • one year ago
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    Having something like this... \[\lim _{x \rightarrow0}e ^{(\frac{ 3 }{ x })\ln(1-3x)}\] is it possible to bring the e out and get this... \[e \lim _{x \rightarrow0}(\frac{ 3 }{ x })\ln (1-3x)\]

  2. SolomonZelman
    • one year ago
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    \(\large\color{black}{ \displaystyle \lim_{x\rightarrow 0}e^{\frac{3}{x}\ln(1-3x)} }\) \(\large\color{black}{ \displaystyle e^{\lim_{x\rightarrow 0}\frac{3}{x}\ln(1-3x)} }\)

  3. SolomonZelman
    • one year ago
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    is this better ?

  4. anonymous
    • one year ago
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    Oh yes! That's it! But what's the logic behind this step? I don't exactly understand

  5. SolomonZelman
    • one year ago
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    you should review the limit properties

  6. anonymous
    • one year ago
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    Hmmm ok gotcha!

  7. SolomonZelman
    • one year ago
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    I mean, nothing offensive, but find a link and read them over. you can find them absolutely everywhere

  8. anonymous
    • one year ago
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    No worries at all! Thanks for the help!

  9. SolomonZelman
    • one year ago
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    \(\LARGE\color{black}{ \displaystyle e^{\lim_{x\rightarrow 0}\frac{3\ln(1-3x)}{x}} }\) \(\LARGE\color{black}{ \displaystyle e^{-\lim_{x\rightarrow 0}\frac{3\ln(1-3x)}{-x}} }\)

  10. SolomonZelman
    • one year ago
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    did this step just now make sense?

  11. SolomonZelman
    • one year ago
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    now both top and bottom go to -infinity

  12. SolomonZelman
    • one year ago
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    go ahead and apply L'H's

  13. anonymous
    • one year ago
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    Is that the same as what you wrote before? (e^ lim....)

  14. SolomonZelman
    • one year ago
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    I didn't change the value. I didn't do any unallowed step

  15. SolomonZelman
    • one year ago
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    am I allowed to multiply times -1 twice? (on the bottom and in front of the lim)

  16. SolomonZelman
    • one year ago
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    \(\LARGE\color{black}{ \displaystyle e^{\lim_{x\rightarrow 0}\frac{3\ln(1-3x)}{x}} }\) \(\LARGE\color{black}{ \displaystyle e^{-\lim_{x\rightarrow 0}\frac{3\ln(1-3x)}{-x}} }\)

  17. SolomonZelman
    • one year ago
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    now apply the lhs to the limit. questions about how I got till there?

  18. anonymous
    • one year ago
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    would we have to multiply by -1 ?

  19. SolomonZelman
    • one year ago
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    what do you mean? where?

  20. SolomonZelman
    • one year ago
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    I already got it to approach -oo

  21. SolomonZelman
    • one year ago
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    to approach -oo on top and bottom

  22. anonymous
    • one year ago
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    Ohhh but it should be approaching 0

  23. SolomonZelman
    • one year ago
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    oh, I mean 0

  24. SolomonZelman
    • one year ago
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    it is.

  25. SolomonZelman
    • one year ago
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    ln(1-0)=1 -x=-0=0

  26. SolomonZelman
    • one year ago
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    why am I saying -oo ? apologize

  27. SolomonZelman
    • one year ago
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    \(\LARGE\color{black}{ \displaystyle e^{-\lim_{x\rightarrow 0}\frac{3\ln(1-3x)}{-x}} }\) top and bottom DO go to 0. ln(1)=0 and -(0)=0

  28. anonymous
    • one year ago
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    So it's just the first line right?

  29. anonymous
    • one year ago
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    I mean I don't get why we'd have a -lim and -x in the denominator

  30. SolomonZelman
    • one year ago
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    \(\LARGE\color{black}{ \displaystyle e^{\lim_{x\rightarrow 0}\frac{3\ln(1-3x)}{x}} }\)

  31. SolomonZelman
    • one year ago
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    it was my mistake, i was thinking of infinity b/c i thought the lim -> (-oo)

  32. SolomonZelman
    • one year ago
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    but in can x->0 take LHS as it is

  33. SolomonZelman
    • one year ago
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    \(\LARGE\color{black}{ \displaystyle e^{\lim_{x\rightarrow 0}\frac{3\ln(1-3x)}{x}} }\) \(\huge \color{black}{ \displaystyle e^{\lim_{x\rightarrow 0}\frac{3\frac{-3}{1-3x}}{1}} }\)

  34. anonymous
    • one year ago
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    Oh okie dokie! So now its just L'Hopital right?

  35. SolomonZelman
    • one year ago
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    yes... that i did d/dx on top and bottm

  36. SolomonZelman
    • one year ago
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    \(\huge \color{black}{ \displaystyle e^{\lim_{x\rightarrow 0}\frac{-9}{1-3x}} }\)

  37. anonymous
    • one year ago
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    I get -9 on the top but on the bottom isn't the derivative of x just 1?

  38. SolomonZelman
    • one year ago
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    yes, 3 replies ago i did the derivative

  39. anonymous
    • one year ago
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    Ohh cause its a fraction over a fraction!

  40. anonymous
    • one year ago
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    Yeah got it!!

  41. SolomonZelman
    • one year ago
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    wel, 1 on bottom for x, and the derivative of ln(1-3x) is a fraction -3/(1-3x)

  42. SolomonZelman
    • one year ago
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    ok, what is your answer?

  43. anonymous
    • one year ago
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    is it e^-9 ? So 1/e^9

  44. SolomonZelman
    • one year ago
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    yes, there you go!

  45. SolomonZelman
    • one year ago
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    I can review the limit properties with you if you want...

  46. anonymous
    • one year ago
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    Finally! Haha! Thanks so much! I've been bugging you a lot today! Im so sorry!

  47. SolomonZelman
    • one year ago
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    Ok, you welcome. You can choose to rvw limit properties later alone, or if you would like to, I can type more here...

  48. anonymous
    • one year ago
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    That's fine! I can look it up on Google real quick! :D

  49. anonymous
    • one year ago
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    Thanks again!

  50. SolomonZelman
    • one year ago
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    Alright. Enjoy:) ... not a problem.

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