Limit involving floor function

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SUMMARY

The limits of the floor function divided by x, specifically lim x→∞ [x]/x and lim x→-∞ [x]/x, both converge to 1. The floor function [x] represents the largest integer not greater than x. The squeeze theorem is applicable, establishing that (x-1)/x < [x]/x ≤ 1 for x > 0. However, for x < 0, the argument requires further examination, as demonstrated by testing x = -1/2.

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  • Understanding of limits in calculus
  • Familiarity with the floor function notation
  • Knowledge of the squeeze theorem
  • Basic algebraic manipulation skills
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Students studying calculus, particularly those focusing on limit evaluation and the properties of the floor function.

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Homework Statement


Evaluate lim x-->infinity [x]/x and lim--> -infinity [x]/x.


Homework Equations





The Attempt at a Solution



The think the limits for both of these are 1. I also know that [x] is the largest integer not greater than x.

I think that I can use the squeeze theorem to say that x-1/x <[x]/x <equal to 1. Does this argument work for both limits or should I prove this an alternate way?
 
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MathSquareRoo said:

Homework Statement


Evaluate lim x-->infinity [x]/x and lim--> -infinity [x]/x.


Homework Equations





The Attempt at a Solution



The think the limits for both of these are 1. I also know that [x] is the largest integer not greater than x.

I think that I can use the squeeze theorem to say that x-1/x <[x]/x <equal to 1. Does this argument work for both limits or should I prove this an alternate way?

For one thing you should use parentheses if you don't want to use TeX. x-1/x is not the same as (x-1)/x and I think you meant (x-1)/x. For another show us how you got that inequality for x>0. And finally you should be able to figure out if it works for x<0. Try x=(-1/2). What went wrong??
 

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