Question

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

Answer 1

ticktock

Answer 2

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

Answer 3

cbrt = cube root

Answer 4

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

Answer 5

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

Answer 6

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

Answer 7

ya thats it

Answer 8

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

Answer 9

i think should be 0

Answer 10

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

Answer 11

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

sorry....

Answer 12

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

Answer 13

well still no limit

Answer 14

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

Answer 15

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

Answer 16

so ultimately its undefined

Answer 17

yes for sure undefined

Answer 18

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

Answer 19

hmm

Answer 20

no

Answer 21

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

Answer 22

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

Answer 23

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

Answer 24

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!

Answer 25

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

Answer 26

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

Answer 27

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

Answer 28

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

Answer 29

ok thanks!