limits and sq root: lim, x-> infty, 8 / sqrt(x) = 1

[bigguy]

OK... i don't know how to start this problem because of the square root.

lim 8 / sqrt(x) = 1
x=> infinite

[nona.m.nona]

OK... i don't know how to start this problem because of the square root.

lim 8 / sqrt(x) = 1
x=> infinite

What are the instructions for this? What are you supposed to be doing with this "limit"? What have you tried? What are your thoughts?

Please be complete. Thank you.

[Martingale]

OK... i don't know how to start this problem because of the square root.

lim 8 / sqrt(x) = 1
x=> infinite

Is this what you mean?...



because it is not true.

[bigguy]

I actually changed some values so the exact problem is not posted publicly...however, the way Martingale displayed it in the post is the way it looks on my paper. Can you tell me step by step on how your coming to the conclusion that its not true?

BTW, if you want to know the real problem, please PM me and I will give you the real problem.

[stapel_eliz]

I actually changed some values so the exact problem is not posted publicly.

Would posting the actual exercise be a violation of some academic-honesty policy? If not, then please post the actual exercise, along with your thoughts and information on your level of study, so the helpers can "see" where you are having trouble. (For instance, did you study horizontal asymptotes back in algebra, but just need a refresher in order to "see" the reasoning behind what has been explained by the helper?)

Thank you!

[Martingale]

I actually changed some values so the exact problem is not posted publicly...however, the way Martingale displayed it in the post is the way it looks on my paper. Can you tell me step by step on how your coming to the conclusion that its not true?

BTW, if you want to know the real problem, please PM me and I will give you the real problem.

when goes to infinity, also goes to infinity.

so for , the numerator is fixed and the denominator is growing without bound. Therefore the limit is ...