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anonymous
 4 years ago
can someone please explain to me why the answer to the following question is infinity, infinity
anonymous
 4 years ago
can someone please explain to me why the answer to the following question is infinity, infinity

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anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[\lim_{x \rightarrow infinity} (\ln(cbrt(x)))/sinx\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0lnx^(1/3)/sinx = (1/3)lnx / sinx ? i thinkk..

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0actually apparently infinity, infinity is not the correct answer.... wolfram lied so if you help me come to the correct answer

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0?? \[\lim_{x\to \infty}\frac{\ln(\sqrt[3]{x})}{\sin(x)}\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0makes no sense since you cannot take the log of a negative number

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0so there is no limit, as the numerator is undefined if \(x\leq 0\)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0oh crap it is \[\lim_{x \rightarrow +infinity}\] sorry....

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\(\sin(x)\) has no limit either as it is periodic and takes on all values between 1 and 1 infinitely often

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0numerator is going to infinity, but denominator is not going to any specific number as i wrote above.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0in other words it swings wildly between large positive and large negative values as sine varies between 1 and 1

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0so ultimately its undefined

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0yes for sure undefined

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0so with that being said, anytime sinx is in my denomonator i know that the function must be undefined?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0well no, the numerator could go to zero, so the whole thing could go to zero

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[\lim_{x\to\infty}\frac{e^{x}}{\sin(x)}\] for example would be zero

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0so are there any good tips and tricks you could tell me to help me understand this sort of material

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0no tricks really, you have to make sure you know what it going on, and don't for example us l'hopital's rule unless it is applicable, i.e. unless you are in indeterminate form, which you are not here. in your example you do not have \(\frac{\infty}{\infty}\) and so you cannot use l\hopital and get zero for an answer!

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0see im not sure how to determine whether i am in that form or not

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0plug in the number and check or imagine what happens as x gets large if you are going to infinity

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0in your example as x gets large so does \(\sqrt[3]{x}\) and therefore so does \(\ln(\sqrt[3]{x})\) but the prolem is that sine does not get large and does not have a limit

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0so basically what i am saying is that in this problem there is no trick or gimmick. you just have to imagine what happens as x gets bigger and bigger
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