Evaluating limit of floor function

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Homework Help Overview

The discussion revolves around evaluating the limit of a function involving the floor function as \( x \) approaches infinity. The original poster presents the limit expression \( \lim_{x\to\infty} \frac{n \log x}{[x]} = 0 \), where \( n \) is a natural number and \( [x] \) denotes the floor function of \( x \).

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants explore the application of the sandwich theorem and discuss the bounds of the floor function. There are attempts to express the floor function in terms of inequalities, such as \( x-1 \leq [x] \leq x \), and to manipulate these inequalities to evaluate the limit.

Discussion Status

Participants have made progress by transforming the inequalities and discussing the implications of the sandwich theorem. There is an ongoing exploration of how the limit behaves as \( x \) approaches infinity, with some noting that \( \frac{\ln x}{x} \) approaches zero.

Contextual Notes

There is a focus on the properties of the floor function and its differentiability, with participants questioning the applicability of L'Hôpital's rule due to the nature of the floor function. The discussion also highlights the need for precise notation regarding the floor and ceiling functions.

Raghav Gupta
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Homework Statement



How $$ \lim_{x\to\infty} \frac{nlogx}{[x]} = 0 $$ ? Here n∈ ℕ
Here [x] is greatest integer or floor function of x.

Homework Equations


[x] = x - {x} where {x} is fractional part of x.

The Attempt at a Solution


I know floor function is not differentiable.
We are getting here ∞/∞ form but can't apply L hopital rule as denominator is not differentiable.
How the limit evaluates to zero?
 
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use sandwich theorem.
hint: start from ##?\le [x]\le ?##
 
AdityaDev said:
use sandwich theorem.
hint: start from ##?\le [x]\le ?##
0≤ {x} < 1
Now what?
 
No. i am asking about [x] and not {x}.
 
Okay,
-∞≤[x] ≤ ∞ but [x] ≠ fractions
 
No. Write it as ##f(x)\le [x]\le g(x)##. what is f and g?
 
AdityaDev said:
No. Write it as ##f(x)\le [x]\le g(x)##. what is f and g?
Why, what is f(x) and g(x) here?
 
Okay. Use ##x-1\le [x]\le x##
 
AdityaDev said:
Okay. Use ##x-1\le [x]\le x##
Okay, what next?
 
  • #10
try to transform that inequality and make it look like ##\frac{lnx}{[x]}## by taking reciprocal and multiplying by lnx. will the inequality change?
 
  • #11
AdityaDev said:
try to transform that inequality and make it look like ##\frac{lnx}{[x]}## by taking reciprocal and multiplying by lnx. will the inequality change?
Then
1/x≤ 1/[x]≤ 1/(x-1)
lnx/x ≤ lnx/[x]≤ lnx/(x-1)
Now?
 
  • #12
Good. Now from sandwich theorem, if you have ##f(x)\le g(x)\le h(x)## and ##\lim_{x \to a}f(x)=L=\lim_{x \to a}h(x)##, then you can say that ##\lim_{x \to a}g(x) = L##.
 
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  • #13
AdityaDev said:
Good. Now from sandwich theorem, if you have ##f(x)\le g(x)\le h(x)## and ##\lim_{x \to a}f(x)=L=\lim_{x \to a}h(x)##, then you can say that ##\lim_{x \to a}g(x) = L##.
Oh, great lnx/x will be zero when x approach infinity same with h(x). So, g(x) evaluates to zero. Thanks.
 
  • #14
Raghav Gupta said:
Oh, great lnx/x will be zero when x approach infinity same with h(x). So, g(x) evaluates to zero. Thanks.
Let's be more precise. ln(x)/x never becomes zero, but gets arbitrarily close to zero as x grows large without bound.
Also, here's the proper notation:
##\lfloor x \rfloor##
\lfloor x \rfloor

For the ceiling function, it is:
##\lceil x \rceil##
\lceil x \rceil
 
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