MHB Limit of integer part function using Sandwich rule

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The limit of the integer part function \(\lim_{x\rightarrow \infty }\frac{[x\cdot a]}{x}\) can be evaluated using the sandwich theorem. By establishing the inequalities \(x\cdot a - 1 < [x\cdot a] < x\cdot a + 1\), it follows that \(\frac{x\cdot a-1}{x}<\frac{[x\cdot a]}{x}< \frac{x\cdot a + 1}{x}\). As \(x\) approaches infinity, the terms \(-\frac{1}{x}\) and \(+\frac{1}{x}\) both converge to 0. Thus, the limit simplifies to the constant \(a\). The conclusion is that the limit is indeed \(a\).
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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