# Limit of a floor function

1. May 5, 2015

### Yoni V

1. The problem statement, all variables and given/known data
The function f is defined f(x)=floor(x^2)/x^2 I need to find the limit of the function at an arbitrary point.

For the continuous parts it was fine, and also for right sided limit at positive points of discontinuity (and left sided for negatives, for all of which the lim is 1), and now I'm left with left sided limit of the function at positive points of discontinuity (and vice versa for the negative part).

I know the answer from intuition: (x2-1)/x2, but I can't find the key to the proof.

2. Relevant equations
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3. The attempt at a solution
Letting the point of discontinuity be x0>0)

First, I restrict δ such that f(x)=(x02-1)/x2

Then, if x0=1 then f(x)-L=0<ε

Otherwise, I restrict my neighborhood again to 1<x<x0

x<x0 ... ->1/x2 - 1/x02>0

and therefore f(x)-L=...=(x02-1)(1/x2 - 1/x02)

And now I'm stuck... I can't seem to find the right combo to make the delta-epsilon magic to work.

Thanks for any help!

2. May 5, 2015

### pasmith

Or, more meaningfully, let the point of discontinuity be $\sqrt{n}$ for strictly positive integer $n$.

Don't use epsilons and deltas unless you have to. Here you do not: it is much easier to use the result that the limit of a product is the product of the limits where the limits exist.

Let $n$ be a strictly positive integer. Then $\lfloor x^2 \rfloor$ is constant on the interval $[\sqrt{n-1},\sqrt{n})$, and $1/x^2$ is continuous at $\sqrt{n}$. What then is $\lim_{x \to \sqrt{n}^{-}} \frac{\lfloor x^2 \rfloor}{x^2}$?

3. May 5, 2015

### Yoni V

Oh my god.... It's so simple... Thanks!
I don't know why it didn't come to mind to use simple arithmetic of limits...
Thanks again