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

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

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

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

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