can someone please explain to me why the answer to the following question is -infinity, infinity

- anonymous

can someone please explain to me why the answer to the following question is -infinity, infinity

- schrodinger

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- anonymous

ticktock

- anonymous

\[\lim_{x \rightarrow -infinity} (\ln(cbrt(x)))/sinx\]

- anonymous

cbrt = cube root

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## More answers

- anonymous

lnx^(1/3)/sinx
= (1/3)lnx / sinx ? i thinkk..

- anonymous

actually apparently -infinity, infinity is not the correct answer.... wolfram lied so if you help me come to the correct answer

- anonymous

??
\[\lim_{x\to -\infty}\frac{\ln(\sqrt[3]{x})}{\sin(x)}\]

- anonymous

ya thats it

- anonymous

makes no sense since you cannot take the log of a negative number

- anonymous

i think should be 0

- anonymous

so there is no limit, as the numerator is undefined if \(x\leq 0\)

- anonymous

oh crap it is \[\lim_{x \rightarrow +infinity}\]
sorry....

- anonymous

\(\sin(x)\) has no limit either as it is periodic and takes on all values between -1 and 1 infinitely often

- anonymous

well still no limit

- anonymous

numerator is going to infinity, but denominator is not going to any specific number as i wrote above.

- anonymous

in other words it swings wildly between large positive and large negative values as sine varies between -1 and 1

- anonymous

so ultimately its undefined

- anonymous

yes for sure undefined

- anonymous

so with that being said, anytime sinx is in my denomonator i know that the function must be undefined?

- anonymous

hmm

- anonymous

no

- anonymous

well no, the numerator could go to zero, so the whole thing could go to zero

- anonymous

\[\lim_{x\to\infty}\frac{e^{-x}}{\sin(x)}\] for example would be zero

- anonymous

so are there any good tips and tricks you could tell me to help me understand this sort of material

- anonymous

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!

- anonymous

see im not sure how to determine whether i am in that form or not

- anonymous

plug in the number and check or imagine what happens as x gets large if you are going to infinity

- anonymous

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

- anonymous

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

- anonymous

ok thanks!

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