Question

Question about limit with floor function:
\[\Large \lim_{x\to\infty}\dfrac{\lfloor\sqrt{x} \rfloor^2}{x}\stackrel{?}{=}1\]

I wasn't sure because there is discontinuous at every square number, but as \(x\) approaches \(\infty\), discontinuous occurs less frequent and eventually disappear. So this limit should be equal to 1. Am I right?

Answer 1

Let's think about this...

Answer 2

The function doesn't necessarily need to be continuous, it just needs to be continuous on a closed interval where the domain approaches.

Answer 3

\[
x>\delta \implies \left|\frac{\lfloor \sqrt x\rfloor ^2}{x}-1\right|<\epsilon
\]

Answer 4

So I'm wondering if the drop caused by the floor function would ever push things out of our epsilon range?

Answer 5

Honestly, I an not familiar with limit definition and I don't see how to apply this in limit where x approaches something such as infinity.

Answer 6

\[
k = \lfloor \sqrt x \rfloor \implies \sqrt{x}-1<k\leq \sqrt x
\]

Answer 7

We can say that \(x>0\) and so \(|x| = x\) because we can pick \(\delta >0\) if we wish.

Answer 8

Hmmm, right now I'm playing around with some ideas here. Suppose: \[
\sqrt x = \lfloor \sqrt{x}\rfloor +a
\]Where \(0\leq a <1\).

Answer 9

Then we can plug it in \(\lfloor \sqrt x \rfloor =\sqrt x-a\). \[
\frac{(\sqrt x - a)^2}{x} = \frac{x^2-2a\sqrt x +a^2}{x}
\]

Answer 10

Whoops.. I messed up on that, didn't I? should be \(x\) not \(x^2\).

Answer 11

\[
1 - \frac{2a\sqrt x}{x} + \frac{a^2}{x}
\]

Answer 12

Wow, now this is clear to me. Smart approach :)