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anonymous

  • one year ago

I'm trying to evaluate this limit using L'Hopital's rule... But apparently I'm doing something wrong! Would appreciate if someone could spot what it is I'm doing wrong!

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  1. anonymous
    • one year ago
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    \[\lim _{x \rightarrow \infty} 6xtan(5/x)\]

  2. anonymous
    • one year ago
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    So I rewrote it as\[\lim _{x \rightarrow \infty} \frac{ \tan(5/x) }{ \frac{ 1 }{ 6x } }\]

  3. anonymous
    • one year ago
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    Once applying L'Hopital's rule I get... \[\frac{ (\frac{ 5 }{ x^2 })\sec^2(5/x) }{ -36x }\]

  4. anonymous
    • one year ago
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    And I believe that the limit of the above equation would be equal to 0... Why is that wrong?

  5. SolomonZelman
    • one year ago
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    can you use the numerical approach. I would do so here.... I am not coming up with a way to re-write this so that you can differentiate top and bottom.... with numerical approach it seems to approach 30 (I plugged in a couple of values). -------------------------------------------------------

  6. anonymous
    • one year ago
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    Hmmm... so just to plug in large numbers?

  7. SolomonZelman
    • one year ago
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    well, I did it for the check. Just that we know the correct answer. I plugged very large numbers to see how the function behaves the more x tends towards infinity.

  8. SolomonZelman
    • one year ago
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    I am trying to think f a good way to this one, if there is.

  9. anonymous
    • one year ago
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    Mhmm right

  10. anonymous
    • one year ago
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    not to butt in but i think your derivative is wrong

  11. anonymous
    • one year ago
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    \[\frac{d}{dx}[\frac{1}{6x}]=-\frac{1}{6x^2}\]

  12. anonymous
    • one year ago
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    also the numerator is off by a minus sign

  13. SolomonZelman
    • one year ago
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    \(\large\color{slate}{\displaystyle\lim_{x \rightarrow ~\infty }~\frac{(-5/x^2)\sec^2\left(\frac{5}{x}\right)}{-\frac{1}{6x^2}}}\) \(\large\color{slate}{\displaystyle\lim_{x \rightarrow ~\infty }~\frac{(5/x^2)\sec^2\left(\frac{5}{x}\right)}{\frac{1}{6x^2}}}\) \(\large\color{slate}{\displaystyle\lim_{x \rightarrow ~\infty }~30\sec^2\left(\frac{5}{x}\right)}\) yes, should be this.

  14. anonymous
    • one year ago
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    that looks better!

  15. SolomonZelman
    • one year ago
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    in my third line, dividing by 1/6 is multiplying times 6. and 1/x^2 cancel

  16. SolomonZelman
    • one year ago
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    yes, now plug in that...

  17. SolomonZelman
    • one year ago
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    30 times (1)^2

  18. anonymous
    • one year ago
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    How does the numerator become negative?

  19. SolomonZelman
    • one year ago
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    because you are taking the derivative of 5/x and that is a negative exponent

  20. anonymous
    • one year ago
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    Oh oh oh!!! Yeah just a confusion sorry!

  21. SolomonZelman
    • one year ago
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    derivat. of (5/x) is the chain rule for your angle....do you get why it is 30 ?

  22. anonymous
    • one year ago
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    Im trying to figure that out...

  23. SolomonZelman
    • one year ago
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    did you get everything up to and including \(\large\color{slate}{\displaystyle\lim_{x \rightarrow ~\infty }~30\sec^2\left(\frac{5}{x}\right)}\) ?

  24. anonymous
    • one year ago
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    Still thinking about the last step

  25. SolomonZelman
    • one year ago
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    ok take your time

  26. anonymous
    • one year ago
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    Ahhh yeah i got it!!

  27. SolomonZelman
    • one year ago
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    yes, so then we have: \(\large\color{slate}{\displaystyle\lim_{x \rightarrow ~\infty }~30\sec^2\left(\frac{5}{x}\right)}\) applying our limit properties, \(\large\color{slate}{\displaystyle 30\sec^2\left(\lim_{x \rightarrow ~\infty }\frac{5}{x}\right)}\)

  28. SolomonZelman
    • one year ago
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    \(\large\color{slate}{\displaystyle 30\sec^2\left(0\right)}\)

  29. anonymous
    • one year ago
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    Thanks so much for the help @SolomonZelman and @satellite73 :)

  30. SolomonZelman
    • one year ago
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    yes, and then you know sec^2(0)=1/cos^2(0)=1/1^2=1

  31. SolomonZelman
    • one year ago
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    so then 30 * 1 ....

  32. anonymous
    • one year ago
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    Yes, I understand the rest :) Thanks so much!

  33. SolomonZelman
    • one year ago
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    yw

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