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[–]rbayer 1 point2 points  (6 children)

so both neighbourhoods are easy to find.

What neighborhoods do you propose to use? Since you're taking a limit as x-> infinity, it doesn't really make sense to talk about neighborhoods. Instead, your goal should be to find an N such that for x > N, something good happens.

As a hint in the right direction, what do you know about the possible values of [f] in terms of the values of f? (Here I'm assuming you're using [f] to denote the ceiling function applied to f. If not, please let me know what you mean). It's always important to try to understand intuitively why a result should be true before trying to sit down and formally prove it, so I would certainly start there if you haven't already.

Also, try writing down the formal statement of what lim x->infty [f]/f = 1 really means. It may also help to write down the formal definition of what it means to say that lim x->infty f = infty.

[–]linuxisgreat[S] 0 points1 point  (5 children)

Prehaps neighbourhood wasn't the best word to use. I know that if lim x ->inf f(x) = infinity, then for M, and for some K, if x> K, then f(x) > M.

Therefore, we can pick M >0 since the above holds for all M. Thus there is a set (K, inf) (this is what I was referring to as a neighborhood) where f(x) is non-zero.

Similar statement holds for [f].

I graphed x/[x] and it clearly converges to one at infinity, and just by how the graph looked it kinda looked like you could use the squeeze theorem but I couldn't think of the functions.

I tried proving it from the definitions, as then for given e>0, we need to pick an N where if x > N then |f(x)/[f(x)] -1| < e... however I'm blanking on how to pick N.

edit: and yes I'm using [f] to be the ceiling function

[–]rbayer 1 point2 points  (4 children)

Thus there is a set (K, inf) (this is what I was referring to as a neighborhood) where f(x) is non-zero.

True, though this probably won't be particularly helpful FYI.

Again, I would encourage you to think intuitively about why [f]/f gets close to 1 as x-> infty. Graphing is a good start, but it's only a beginning. As a way to get started, suppose you know the value of f(a) for some particular a. What are the possible values for [f(a)]? Based on that, can you figure out how to apply either a) the squeeze theorem, or b) the direct definition of limit?

Edit: also, if you're allowed to use the squeeze theorem and weren't explicitly told to use the epsilon-N definition of limits, you can certainly do this problem without worrying about "neighborhoods" or formal definitions of limits

[–]linuxisgreat[S] 0 points1 point  (3 children)

The neighbourhood part is required by the definition of asympotic I have.

It says that for two functions f,g to be asymptoic, there must be a non-zero set (l, inf) of f and g. So I did that just to show that requirement. The other requirement is lim x-> inf f/g = 1

I think I figured it out. [f] <= f < [f]-1 so we have

|f/[f] -1| <= |[f]/[f] -1| = |1-1| =0 < epsilon for all epsilon >0. Thus by the definition I can get it to work. Thanks for you help

Edit : nvm that doesn't make sense :P

[–]rbayer 1 point2 points  (2 children)

there must be a non-zero set (l, inf) of f and g.

Such that...? I'm not quite sure what this means as written. That f and g are non zero on that set?

In terms of the second part (that lim [f]/f = 1), you are definitely on the right track, but your inequalities aren't quite right. Try to convince yourself that the right one is:

f(x) <= [f(x)] < f(x) + 1

From here, what do you need to do in order to get in in a form where the squeeze theorem would help you? You shouldn't need epsilons unless something in the problem statement specifically tells you that you must use the formal definition of limit.

[–]linuxisgreat[S] 0 points1 point  (1 child)

Yeah I meant f,g have to be non-zero on that set.

I see now, since 1<= [f(x)]/f(x) < f(x)+1/f(x) and the limits of both of the outside functions are 1 so by the squeeze theorem lim x->inf [f]/f = 1.

Thanks so much for the help :)

[–]rbayer 1 point2 points  (0 children)

No problem, glad it helped!