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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!
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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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[\lim _{x \rightarrow \infty} 6xtan(5/x)\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0So I rewrote it as\[\lim _{x \rightarrow \infty} \frac{ \tan(5/x) }{ \frac{ 1 }{ 6x } }\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Once applying L'Hopital's rule I get... \[\frac{ (\frac{ 5 }{ x^2 })\sec^2(5/x) }{ 36x }\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0And I believe that the limit of the above equation would be equal to 0... Why is that wrong?

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.1can you use the numerical approach. I would do so here.... I am not coming up with a way to rewrite 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
 one year ago
Best ResponseYou've already chosen the best response.0Hmmm... so just to plug in large numbers?

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.1well, 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
 one year ago
Best ResponseYou've already chosen the best response.1I am trying to think f a good way to this one, if there is.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0not to butt in but i think your derivative is wrong

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[\frac{d}{dx}[\frac{1}{6x}]=\frac{1}{6x^2}\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0also the numerator is off by a minus sign

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.1\(\large\color{slate}{\displaystyle\lim_{x \rightarrow ~\infty }~\frac{(5/x^2)\sec^2\left(\frac{5}{x}\right)}{\frac{1}{6x^2}}}\) \(\large\color{slate}{\displaystyle\lim_{x \rightarrow ~\infty }~\frac{(5/x^2)\sec^2\left(\frac{5}{x}\right)}{\frac{1}{6x^2}}}\) \(\large\color{slate}{\displaystyle\lim_{x \rightarrow ~\infty }~30\sec^2\left(\frac{5}{x}\right)}\) yes, should be this.

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.1in my third line, dividing by 1/6 is multiplying times 6. and 1/x^2 cancel

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.1yes, now plug in that...

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0How does the numerator become negative?

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.1because you are taking the derivative of 5/x and that is a negative exponent

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Oh oh oh!!! Yeah just a confusion sorry!

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.1derivat. of (5/x) is the chain rule for your angle....do you get why it is 30 ?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Im trying to figure that out...

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.1did you get everything up to and including \(\large\color{slate}{\displaystyle\lim_{x \rightarrow ~\infty }~30\sec^2\left(\frac{5}{x}\right)}\) ?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Still thinking about the last step

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.1ok take your time

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Ahhh yeah i got it!!

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.1yes, 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
 one year ago
Best ResponseYou've already chosen the best response.1\(\large\color{slate}{\displaystyle 30\sec^2\left(0\right)}\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Thanks so much for the help @SolomonZelman and @satellite73 :)

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.1yes, and then you know sec^2(0)=1/cos^2(0)=1/1^2=1

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.1so then 30 * 1 ....

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Yes, I understand the rest :) Thanks so much!
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