Limiting x->0: Analyzing f(x) = 1/|[[x]]|

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Homework Statement


Limit x->0 f(x)
where f(x) = 1/|[[x]]|
where [[x]] is the largest integer function

Homework Equations





The Attempt at a Solution


I am having problem redefining the absolute value here
|[[x]]| = -[[x]] when [[x]] <0
so that is when x<0?
the right hand limit is supposed to be infinity
but I am confused about the left one is it 1?
or is it negative infinity or infinity?
 
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You are overthinking this. To take the "limit as x goes to 0", look at points close to x= 0. What is [[0.00001]]? What is [[-0.000001]]? What are the absolute values of those?

Do "limit from the right" (x> 0) and "limit from the left" (x< 0) separately.
 
it isn't the absolute value function its the greatest integer function
http://www.icoachmath.com/SiteMap/GreatestIntegerFunction.html
but my problem is that since .9999999 continuously is 1 should I consider the greatest integer of -.00000000000...01 as 0 or -1?
 
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You said it was "the absolute value of the greatest integer function".

And the greatest integer less than or equal to 0.0000000...01 (where the ... must mean some specific number of missing 0s, not an infinite string) is 0, of course. Since it is positive why would you even consider -1? That has nothing to do with "0.9999999..." being equal to 1.
 
the limit from the right is 0 and the left its 1 right?
 
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Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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