# Thread: Asymptotic behavior of the number of perfect squares less than x

1. ## Asymptotic behavior of the number of perfect squares less than x

Let $S_x=|\{ n \le x : \left(\exists m \right) \left( n=m^2\right), n\in \mathbb{N} \}|$. Find $\mathop {\lim }\limits_{x \to + \infty } \frac{{S_x}}{{x}}$.

I'm thinking that $S_x$ would be the greatest $m\in\mathbb{N}$ such that $m \le \sqrt{x}$ plus 1 or $S_x=\left[\sqrt {x} \right]+1$.

The limit would then be $\mathop {\lim }\limits_{x \to + \infty } \frac{{S_x}}{{x}}=\mathop {\lim }\limits_{x \to + \infty } \frac{{\left[\sqrt {x} \right]+1}}{{x}}$.

I'm not sure if the argument is valid but if it is, the problem I have is finding $\left[\sqrt{x}\right]+1$ for cases of $x$. Any help is appreciated.

2. Your reasoning is perfectly valid. The percentage of square numbers less than or equal to x is approximately $\sqrt(x)$ in x which tends to zero as x tends to infinity.