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kaiz122 Group Title

help me here. please

  • one year ago
  • one year ago

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  1. kaiz122 Group Title
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    \[\lim_{x \rightarrow +\infty} (1+ \frac{k}{x})^{x}\]

    • one year ago
  2. satellite73 Group Title
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    any guesses?

    • one year ago
  3. kaiz122 Group Title
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    it's in indeterminate form \[1^{\infty}\]

    • one year ago
  4. satellite73 Group Title
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    since on definition of \(e\) is \(\lim_{x\to \infty}(1+\frac{1}{x})^x\) it will be no surprise that \[\lim_{x\to \infty}\left(1+\frac{k}{x}\right)^x=e^k\]

    • one year ago
  5. satellite73 Group Title
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    if that is an unacceptable answer, take the log, take the limit, which will be \(k\) and then exponentiate to get \(e^k\)

    • one year ago
  6. satellite73 Group Title
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    start with \[x\ln(1+\frac{1}{x})\] which is now \(\infty\times 0\) then the usual trick is to rewrite as \[\frac{\ln(1+\frac{1}{x})}{\frac{1}{x}}\] so it is now \(\frac{0}{0}\) and then use l'hopital

    • one year ago
  7. kaiz122 Group Title
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    let \[y=(1+\frac{k}{x})^{x}\] \[\ln{y} =x\ln(1+\frac{k}{x})\] then, \[\lim_{x \rightarrow +\infty} x \ln (1+\frac{k}{x})\]

    • one year ago
  8. satellite73 Group Title
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    yup then go to \[\frac{\ln(1+\frac{k}{x})}{\frac{1}{x}}\] and you should be good to go

    • one year ago
  9. kaiz122 Group Title
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    =0 *(+Infinity)

    • one year ago
  10. satellite73 Group Title
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    right rewrite as i did above to get \(\frac{0}{0}\)

    • one year ago
  11. kaiz122 Group Title
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    we can rewrite it to this form \[\lim_{x \rightarrow +\infty} \frac{\ln(1+ \frac{k}{x})}{\frac{1}{x}}\]

    • one year ago
  12. kaiz122 Group Title
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    =\[\frac{0}{0}\]

    • one year ago
  13. kaiz122 Group Title
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    should we use the L' Hospital Rule now?

    • one year ago
  14. satellite73 Group Title
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    yes

    • one year ago
  15. kaiz122 Group Title
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    what's the derivative of \[\ln (1+\frac{k}{x})?\] is it \[\frac{x^{2}}{(x+k)^{2}}?\]

    • one year ago
  16. phi Group Title
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    no.

    • one year ago
  17. kaiz122 Group Title
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    ok so it's \[\frac{-k}{x^{2}+kx}\]

    • one year ago
  18. sunnymony Group Title
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    u know 1^infinite will be =o

    • one year ago
  19. kaiz122 Group Title
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    i don't think it is.

    • one year ago
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