Open study

is now brainly

With Brainly you can:

  • Get homework help from millions of students and moderators
  • Learn how to solve problems with step-by-step explanations
  • Share your knowledge and earn points by helping other students
  • Learn anywhere, anytime with the Brainly app!

A community for students.

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

See more answers at
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly


Get your free account and access expert answers to this and thousands of other questions

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

Not the answer you are looking for?

Search for more explanations.

Ask your own question

Other answers:

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

Not the answer you are looking for?

Search for more explanations.

Ask your own question