Limit of integer part function using Sandwich rule

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Discussion Overview

The discussion centers around calculating the limit of the integer part function using the Sandwich rule, specifically the limit \(\lim_{x\rightarrow \infty }\frac{[x\cdot a]}{x}\). Participants explore the properties of the integer part function and its implications for the limit as \(x\) approaches infinity.

Discussion Character

  • Exploratory, Mathematical reasoning

Main Points Raised

  • One participant introduces the limit calculation and seeks assistance on how to approach it.
  • Another participant provides a mathematical framework using the properties of the integer part function, stating that \(u-1 < [u] < u+1\) and applying it to the expression \(\frac{[x\cdot a]}{x}\).
  • This participant derives inequalities that lead to the conclusion that \(\frac{[x\cdot a]}{x}\) is bounded between \(a - \frac{1}{x}\) and \(a + \frac{1}{x}\) as \(x\) approaches infinity.
  • Subsequent replies express agreement with the derived conclusion that the limit approaches the constant \(a\) as \(x\) goes to infinity, noting that \(\frac{1}{x}\) approaches 0.

Areas of Agreement / Disagreement

Participants appear to agree on the approach and the conclusion that the limit is the constant \(a\) as \(x\) approaches infinity, although the initial query remains open to further exploration.

Contextual Notes

The discussion relies on the properties of the integer part function and the application of the Sandwich theorem, with assumptions about the behavior of \(x\) as it approaches infinity. The implications of the integer part function's rounding behavior are central to the reasoning presented.

Yankel
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Hello everyone,

I want to calculate the following limits:

\[\lim_{x\rightarrow \infty }\frac{[x\cdot a]}{x}\]

using the sandwich rule, where [xa] is the integer part function defined here:

Integer Part -- from Wolfram MathWorld

I am not sure how to approach this. Any assistance will be most appreciated.
 
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For the integer part of some $u$, we have:
$$u-1 < < u+1$$
That is, we are rounding either up or down, meaning that we end up less than 1 higher or lower.

So:
$$ x\cdot a - 1 < [x\cdot a] < x\cdot a + 1$$
With positive $x$, it follows that:
$$\frac{x\cdot a-1}{x}<\frac{[x\cdot a]}{x}< \frac{x\cdot a + 1}{x} \implies a-\frac 1x < \frac{[x\cdot a]}{x} < a + \frac 1x$$
Now let $x$ go to infinity and apply the sandwich theorem.
 
Great idea...

so the correct answer is the constant a then ?

If x goes to infinity, 1/x goes to 0
 
Yankel said:
Great idea...

so the correct answer is the constant a then ?

If x goes to infinity, 1/x goes to 0

Yep. (Nod)
 

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