anonymous
  • anonymous
can someone please explain to me why the answer to the following question is -infinity, infinity
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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anonymous
  • anonymous
ticktock
anonymous
  • anonymous
\[\lim_{x \rightarrow -infinity} (\ln(cbrt(x)))/sinx\]
anonymous
  • anonymous
cbrt = cube root

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anonymous
  • anonymous
lnx^(1/3)/sinx = (1/3)lnx / sinx ? i thinkk..
anonymous
  • anonymous
actually apparently -infinity, infinity is not the correct answer.... wolfram lied so if you help me come to the correct answer
anonymous
  • anonymous
?? \[\lim_{x\to -\infty}\frac{\ln(\sqrt[3]{x})}{\sin(x)}\]
anonymous
  • anonymous
ya thats it
anonymous
  • anonymous
makes no sense since you cannot take the log of a negative number
anonymous
  • anonymous
i think should be 0
anonymous
  • anonymous
so there is no limit, as the numerator is undefined if \(x\leq 0\)
anonymous
  • anonymous
oh crap it is \[\lim_{x \rightarrow +infinity}\] sorry....
anonymous
  • anonymous
\(\sin(x)\) has no limit either as it is periodic and takes on all values between -1 and 1 infinitely often
anonymous
  • anonymous
well still no limit
anonymous
  • anonymous
numerator is going to infinity, but denominator is not going to any specific number as i wrote above.
anonymous
  • anonymous
in other words it swings wildly between large positive and large negative values as sine varies between -1 and 1
anonymous
  • anonymous
so ultimately its undefined
anonymous
  • anonymous
yes for sure undefined
anonymous
  • anonymous
so with that being said, anytime sinx is in my denomonator i know that the function must be undefined?
anonymous
  • anonymous
hmm
anonymous
  • anonymous
no
anonymous
  • anonymous
well no, the numerator could go to zero, so the whole thing could go to zero
anonymous
  • anonymous
\[\lim_{x\to\infty}\frac{e^{-x}}{\sin(x)}\] for example would be zero
anonymous
  • anonymous
so are there any good tips and tricks you could tell me to help me understand this sort of material
anonymous
  • 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
  • anonymous
see im not sure how to determine whether i am in that form or not
anonymous
  • anonymous
plug in the number and check or imagine what happens as x gets large if you are going to infinity
anonymous
  • 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
  • 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
  • anonymous
ok thanks!

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