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kaiz122

  • 3 years ago

help me here. please

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  1. kaiz122
    • 3 years ago
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    \[\lim_{x \rightarrow +\infty} (1+ \frac{k}{x})^{x}\]

  2. anonymous
    • 3 years ago
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    any guesses?

  3. kaiz122
    • 3 years ago
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    it's in indeterminate form \[1^{\infty}\]

  4. anonymous
    • 3 years ago
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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 \[\lim_{x\to \infty}\left(1+\frac{k}{x}\right)^x=e^k\]

  5. anonymous
    • 3 years ago
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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 \(e^k\)

  6. anonymous
    • 3 years ago
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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

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

  8. anonymous
    • 3 years ago
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    yup then go to \[\frac{\ln(1+\frac{k}{x})}{\frac{1}{x}}\] and you should be good to go

  9. kaiz122
    • 3 years ago
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    =0 *(+Infinity)

  10. anonymous
    • 3 years ago
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    right rewrite as i did above to get \(\frac{0}{0}\)

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

  12. kaiz122
    • 3 years ago
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    =\[\frac{0}{0}\]

  13. kaiz122
    • 3 years ago
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    should we use the L' Hospital Rule now?

  14. anonymous
    • 3 years ago
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    yes

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

  16. phi
    • 3 years ago
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    no.

  17. kaiz122
    • 3 years ago
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    ok so it's \[\frac{-k}{x^{2}+kx}\]

  18. sunnymony
    • 3 years ago
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    u know 1^infinite will be =o

  19. kaiz122
    • 3 years ago
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    i don't think it is.

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