anonymous
  • anonymous
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!
Mathematics
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anonymous
  • anonymous
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!
Mathematics
schrodinger
  • schrodinger
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anonymous
  • anonymous
\[\lim _{x \rightarrow \infty} 6xtan(5/x)\]
anonymous
  • anonymous
So I rewrote it as\[\lim _{x \rightarrow \infty} \frac{ \tan(5/x) }{ \frac{ 1 }{ 6x } }\]
anonymous
  • anonymous
Once applying L'Hopital's rule I get... \[\frac{ (\frac{ 5 }{ x^2 })\sec^2(5/x) }{ -36x }\]

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anonymous
  • anonymous
And I believe that the limit of the above equation would be equal to 0... Why is that wrong?
SolomonZelman
  • SolomonZelman
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). -------------------------------------------------------
anonymous
  • anonymous
Hmmm... so just to plug in large numbers?
SolomonZelman
  • SolomonZelman
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.
SolomonZelman
  • SolomonZelman
I am trying to think f a good way to this one, if there is.
anonymous
  • anonymous
Mhmm right
anonymous
  • anonymous
not to butt in but i think your derivative is wrong
anonymous
  • anonymous
\[\frac{d}{dx}[\frac{1}{6x}]=-\frac{1}{6x^2}\]
anonymous
  • anonymous
also the numerator is off by a minus sign
SolomonZelman
  • SolomonZelman
\(\large\color{slate}{\displaystyle\lim_{x \rightarrow ~\infty }~\frac{(-5/x^2)\sec^2\left(\frac{5}{x}\right)}{-\frac{1}{6x^2}}}\) \(\large\color{slate}{\displaystyle\lim_{x \rightarrow ~\infty }~\frac{(5/x^2)\sec^2\left(\frac{5}{x}\right)}{\frac{1}{6x^2}}}\) \(\large\color{slate}{\displaystyle\lim_{x \rightarrow ~\infty }~30\sec^2\left(\frac{5}{x}\right)}\) yes, should be this.
anonymous
  • anonymous
that looks better!
SolomonZelman
  • SolomonZelman
in my third line, dividing by 1/6 is multiplying times 6. and 1/x^2 cancel
SolomonZelman
  • SolomonZelman
yes, now plug in that...
SolomonZelman
  • SolomonZelman
30 times (1)^2
anonymous
  • anonymous
How does the numerator become negative?
SolomonZelman
  • SolomonZelman
because you are taking the derivative of 5/x and that is a negative exponent
anonymous
  • anonymous
Oh oh oh!!! Yeah just a confusion sorry!
SolomonZelman
  • SolomonZelman
derivat. of (5/x) is the chain rule for your angle....do you get why it is 30 ?
anonymous
  • anonymous
Im trying to figure that out...
SolomonZelman
  • SolomonZelman
did you get everything up to and including \(\large\color{slate}{\displaystyle\lim_{x \rightarrow ~\infty }~30\sec^2\left(\frac{5}{x}\right)}\) ?
anonymous
  • anonymous
Still thinking about the last step
SolomonZelman
  • SolomonZelman
ok take your time
anonymous
  • anonymous
Ahhh yeah i got it!!
SolomonZelman
  • SolomonZelman
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, \(\large\color{slate}{\displaystyle 30\sec^2\left(\lim_{x \rightarrow ~\infty }\frac{5}{x}\right)}\)
SolomonZelman
  • SolomonZelman
\(\large\color{slate}{\displaystyle 30\sec^2\left(0\right)}\)
anonymous
  • anonymous
Thanks so much for the help @SolomonZelman and @satellite73 :)
SolomonZelman
  • SolomonZelman
yes, and then you know sec^2(0)=1/cos^2(0)=1/1^2=1
SolomonZelman
  • SolomonZelman
so then 30 * 1 ....
anonymous
  • anonymous
Yes, I understand the rest :) Thanks so much!
SolomonZelman
  • SolomonZelman
yw

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