Limit of integer part function using Sandwich rule

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SUMMARY

The limit of the integer part function using the Sandwich Rule is calculated as follows: \[\lim_{x\rightarrow \infty }\frac{[x\cdot a]}{x} = a\]. The integer part function, denoted as \([x\cdot a]\), satisfies the inequalities \(x\cdot a - 1 < [x\cdot a] < x\cdot a + 1\). By applying the Sandwich Theorem, as \(x\) approaches infinity, the terms \(-\frac{1}{x}\) and \(+\frac{1}{x}\) converge to zero, confirming that the limit is indeed the constant \(a\).

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  • Understanding of limits in calculus
  • Familiarity with the integer part function
  • Knowledge of the Sandwich Theorem
  • Basic algebraic manipulation skills
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  • Study the properties of the integer part function in detail
  • Explore advanced applications of the Sandwich Theorem
  • Learn about limits involving piecewise functions
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Students and professionals in mathematics, particularly those studying calculus and real analysis, as well as educators seeking to explain the concept of limits and the integer part function.

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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