Limits of Function f(x) with Greatest Integer [x]: Problem Analysis

[abcd8989]

Given a function f(x) = lim n->infintiy [x]^n / (x^n +1) , where [x] is a greatest integer function.
What is the limit value of lim x->1+ f(x) ?
Is the limit found above the same with lim n->infinity (lim x->1+ [x]^n / (x^n +1) ) ?
I am rather confused with the above two cases. I don't know how to think of it. I have such kind of thought: if n tends to infinity first, even x becomes very close to 1, the value is still at infinity. If x tends to 1 faster, even n tends to infinity, the value is still 1.
What's wrong with my concepts?

 

[Dick]

So [x] means the greatest integer less than or equal to x? If so, what is the limit x->1+ of [x]/(1+x^n)?

 

[abcd8989]

Yes, but I m not sure about the limit..I think it should be 1..

 

[Char. Limit]

Note that both numerator and denominator exist and are nonzero for all x>1, and thus the limit is rather easy to evaluate.

 

[icesalmon]

as x increases without bound, the greatest integer returns decimal values of n.something back to n, as the denominator keeps growing; approaching infinity.

 

[HallsofIvy]

In fact, since you want to take the limit as x goes to 1 from above, you only need to look at values of x between 1 and 2.

1^n is pretty easy, isn't it?

 

[Char. Limit]

For that matter, try just plugging 1 into the equation.