## m.auld64 4 years ago can someone please explain to me why the answer to the following question is -infinity, infinity

1. jaersyn

ticktock

2. m.auld64

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

3. m.auld64

cbrt = cube root

4. jaersyn

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

5. m.auld64

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

6. anonymous

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

7. m.auld64

ya thats it

8. anonymous

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

9. myko

i think should be 0

10. anonymous

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

11. m.auld64

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

12. anonymous

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

13. anonymous

well still no limit

14. anonymous

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

15. anonymous

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

16. m.auld64

so ultimately its undefined

17. anonymous

yes for sure undefined

18. m.auld64

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

19. myko

hmm

20. jaersyn

no

21. anonymous

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

22. anonymous

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

23. m.auld64

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

24. 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!

25. m.auld64

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

26. anonymous

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

27. 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

28. 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

29. m.auld64

ok thanks!