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Dido525

  • 3 years ago

Find the Limit:

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  1. Dido525
    • 3 years ago
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    \[\lim_{x \rightarrow \infty}(\frac{ 5x-3 }{ 5x+7 })^{3x+1}\]

  2. anonymous
    • 3 years ago
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    take the log take the limit exponentiate

  3. Dido525
    • 3 years ago
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    I am allowed to use L'Hospital's rule. I am not sure how I would use it though...

  4. Dido525
    • 3 years ago
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    So I get it's (infinity/infinity)^(infinity) but I am not sure how to go from there...

  5. Dido525
    • 3 years ago
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    Yep... I am not sure how that would help though...

  6. anonymous
    • 3 years ago
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    \[(3x+1)\ln(\frac{5x-3}{5x+7})\] now you \(\infty\times 0\)

  7. Dido525
    • 3 years ago
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    Yep... I got there.

  8. anonymous
    • 3 years ago
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    rewrite as \[\frac{\ln(\frac{5x-3}{5x+7})}{\frac{1}{3x+1}}\] now you have \[\frac{0}{0}\] and now you can use l'hopital

  9. Dido525
    • 3 years ago
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    Hmm Okay... Let's see...

  10. anonymous
    • 3 years ago
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    probably easier if you write \[\frac{\ln(5x-3)-\ln(5x+7)}{\frac{1}{3x+1}}\]

  11. Dido525
    • 3 years ago
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    Okay Thanks!

  12. Dido525
    • 3 years ago
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    Yep. I know.

  13. anonymous
    • 3 years ago
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    yw

  14. Dido525
    • 3 years ago
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    @satellite73 : Okay so I have: \[\lim_{x \rightarrow \infty} \frac{ \frac{ 5 }{ 5x-3 }-\frac{ 5 }{ 5x+7 } }{ \frac{ -2 }{ (3x+1)^2 } }\]

  15. anonymous
    • 3 years ago
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    denominator is wrong

  16. anonymous
    • 3 years ago
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    \[\lim_{x \rightarrow \infty} \frac{ \frac{ 5 }{ 5x-3 }-\frac{ 5 }{ 5x+7 } }{ \frac{ -3 }{ (3x+1)^2 } }\]

  17. Dido525
    • 3 years ago
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    Chain rule. Right.

  18. anonymous
    • 3 years ago
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    ugly retricealgebra from here on in have fun, but you will get a rational expression so limit is same as horizontal asymptote

  19. Dido525
    • 3 years ago
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    :( . Fair enough....

  20. anonymous
    • 3 years ago
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    actually it is not that bad

  21. Dido525
    • 3 years ago
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    Yep. I think I see why.

  22. anonymous
    • 3 years ago
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    \[-\frac{(3x+1)^2}{3}\times \frac{50}{(5x-3)(5x+7)}\] and not need to multiply out, just look at the ratio of the leading coefficients

  23. anonymous
    • 3 years ago
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    i mean you have to be a bit careful but the leading coefficient of the numerator is \(9\times 50\) and the leading coefficient of the denominator is \(-3\times 25\) so you the ratio is \(-6\)

  24. anonymous
    • 3 years ago
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    and don't forget to exponentiate at the end

  25. Dido525
    • 3 years ago
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    Hmm... Okay.

  26. Dido525
    • 3 years ago
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    So I got 1/e^6 or e^-6.

  27. anonymous
    • 3 years ago
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    yup

  28. Dido525
    • 3 years ago
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    :) . Thanks a million :) .

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