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Question Number 8785 by FilupSmith last updated on 27/Oct/16

Commented by FilupSmith last updated on 27/Oct/16

$$f\left(x\right)=\sqrt{x^2-\lfloor x\rfloor }$$$$$$$$x=N+c,\lfloor x\rfloor =N\in \mathbb{Z}$$$$0⩽c<1$$$$\mathrm{for}-1<x<1\Rightarrow x^2-\lfloor x^2\rfloor ={(0+c)}^2+\lfloor {(0+c)}^2\rfloor$$$$=c^2(\sqrt{x^2-\lfloor x^2\rfloor }=\mid c\mid )$$$$\uparrow \mathrm{explains}\mathrm{the}\mathrm{straight}\mathrm{line}$$$$$$$$\mathrm{Question}:$$$$\mathrm{Why}\mathrm{do}\mathrm{the}\mathrm{curved}\mathrm{lines}\mathrm{get}\mathrm{steaper}$$$$\mathrm{and}\mathrm{closer}\mathrm{together}?$$

Commented by 123456 last updated on 27/Oct/16

$$t=\sqrt{n}$$$$\mathrm{\Delta }t=\sqrt{n+1}-\sqrt{n}$$$$\underset{n\to \mathrm{\infty }}{\lim }\sqrt{n+1}-\sqrt{n}$$$$=\underset{n\to \mathrm{\infty }}{\lim }\frac{(\sqrt{n+1}-\sqrt{n})(\sqrt{n+1}+\sqrt{n})}{\sqrt{n+1}+\sqrt{n}}$$$$=\underset{n\to \mathrm{\infty }}{\lim }\frac{1}{\sqrt{n+1}+\sqrt{n}}\to \frac{1}{\mathrm{\infty }}$$$$=0$$

Commented by FilupSmith last updated on 28/Oct/16

$$\mathrm{Thank}\mathrm{you},\mathrm{this}\mathrm{helps}\mathrm{me}\mathrm{understand}$$

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