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Question Number 62747 by Mikael last updated on 24/Jun/19

$$Are\mathit{f},\mathit{g}:\mathbb{R}\to \mathbb{R}definedby$$$$f\left(x\right)=\{\begin{array}{l}0,x\in \mathbb{R}\mathrm{\setminus }\mathbb{Q}\\ x,x\in \mathbb{Q}\end{array}$$$$g\left(x\right)=\{\begin{array}{l}1,x=0\\ 0,x\ne 0\end{array}$$$$showthat\underset{\mathbf{x}\to 0}{\mathbf{lim}}\mathbf{f}\left(\mathbf{x}\right)=0and\underset{\mathbf{y}\to 0}{\mathbf{lim}}\mathbf{g}\left(\mathbf{y}\right)=0$$$$however\underset{\mathbf{x}\to 0}{\mathbf{lim}}\mathbf{g}\left(\mathbf{f}\left(\mathbf{x}\right)\right)doesnotexist.$$

Answered by MJS last updated on 25/Jun/19

$$x\in \mathbb{R}\mathrm{\setminus }\mathbb{Q}:\underset{x\to 0}{\lim }f\left(x\right)=0\mathrm{is}\mathrm{trivial}\mathrm{because}f\left(x\right)=0\mathrm{in}\mathrm{this}\mathrm{case}$$$$x\in \mathbb{Q}:\underset{x\to 0}{\lim }f\left(x\right)=\underset{p\to 0}{\lim }\frac{p}{q}\mathrm{\forall }p\in \mathbb{Z}\wedge \mathrm{\forall }q\in {\mathbb{N}}^{★}\Rightarrow$$$$\Rightarrow \underset{x\to 0}{\lim }f\left(x\right)=\underset{p\to 0}{\lim }\frac{p}{q}=0$$$$\underset{x\to 0}{\lim }g\left(x\right)=0\mathrm{because}g\left(x\right)=0\mathrm{for}x\in [-ϵ;ϵ]\mathrm{\forall }ϵ\in {\mathbb{R}}^{+}$$$$\underset{x\to 0}{\lim }g\left(f\left(x\right)\right)\mathrm{does}\mathrm{not}\mathrm{exist}\mathrm{because}\mathrm{\forall }x\in \mathbb{R}\mathrm{\setminus }\mathbb{Q}$$$$g\left(f\left(x\right)\right)=1\mathrm{but}\mathrm{\forall }x\in \mathbb{Q}g\left(f\left(x\right)\right)=0\Rightarrow$$$$\Rightarrow \underset{x\in \mathbb{R}\mathrm{\setminus }\mathbb{Q}\to 0}{\lim }g\left(f\left(x\right)\right)=1\wedge \underset{x\in \mathbb{Q}\to 0}{\lim }g\left(f\left(x\right)\right)=0\Rightarrow$$$$\mathrm{limit}\mathrm{\forall }x\in \mathbb{R}{\mathrm{doesn}}^{\prime }\mathrm{t}\mathrm{exist}$$

Commented by Mikael last updated on 25/Jun/19

$$thankyouSir.$$

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