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

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m.auld64Best ResponseYou've already chosen the best response.0
\[\lim_{x \rightarrow infinity} (\ln(cbrt(x)))/sinx\]
 2 years ago

jaersynBest ResponseYou've already chosen the best response.1
lnx^(1/3)/sinx = (1/3)lnx / sinx ? i thinkk..
 2 years ago

m.auld64Best ResponseYou've already chosen the best response.0
actually apparently infinity, infinity is not the correct answer.... wolfram lied so if you help me come to the correct answer
 2 years ago

satellite73Best ResponseYou've already chosen the best response.0
?? \[\lim_{x\to \infty}\frac{\ln(\sqrt[3]{x})}{\sin(x)}\]
 2 years ago

satellite73Best ResponseYou've already chosen the best response.0
makes no sense since you cannot take the log of a negative number
 2 years ago

satellite73Best ResponseYou've already chosen the best response.0
so there is no limit, as the numerator is undefined if \(x\leq 0\)
 2 years ago

m.auld64Best ResponseYou've already chosen the best response.0
oh crap it is \[\lim_{x \rightarrow +infinity}\] sorry....
 2 years ago

satellite73Best 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
 2 years ago

satellite73Best ResponseYou've already chosen the best response.0
well still no limit
 2 years ago

satellite73Best ResponseYou've already chosen the best response.0
numerator is going to infinity, but denominator is not going to any specific number as i wrote above.
 2 years ago

satellite73Best ResponseYou've already chosen the best response.0
in other words it swings wildly between large positive and large negative values as sine varies between 1 and 1
 2 years ago

m.auld64Best ResponseYou've already chosen the best response.0
so ultimately its undefined
 2 years ago

satellite73Best ResponseYou've already chosen the best response.0
yes for sure undefined
 2 years ago

m.auld64Best ResponseYou've already chosen the best response.0
so with that being said, anytime sinx is in my denomonator i know that the function must be undefined?
 2 years ago

satellite73Best ResponseYou've already chosen the best response.0
well no, the numerator could go to zero, so the whole thing could go to zero
 2 years ago

satellite73Best ResponseYou've already chosen the best response.0
\[\lim_{x\to\infty}\frac{e^{x}}{\sin(x)}\] for example would be zero
 2 years ago

m.auld64Best ResponseYou've already chosen the best response.0
so are there any good tips and tricks you could tell me to help me understand this sort of material
 2 years ago

satellite73Best ResponseYou've already chosen the best response.0
no 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!
 2 years ago

m.auld64Best ResponseYou've already chosen the best response.0
see im not sure how to determine whether i am in that form or not
 2 years ago

satellite73Best ResponseYou've already chosen the best response.0
plug in the number and check or imagine what happens as x gets large if you are going to infinity
 2 years ago

satellite73Best ResponseYou've already chosen the best response.0
in 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
 2 years ago

satellite73Best ResponseYou've already chosen the best response.0
so 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
 2 years ago
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