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Question Number 165441 by mnjuly1970 last updated on 01/Feb/22

Answered by mahdipoor last updated on 01/Feb/22

$$\mathrm{if}\mathrm{g}\left(\mathrm{x}\right)\mathrm{is}\mathrm{continuous}\mathrm{and}\mathrm{increment}$$$$\mathrm{function}\mathrm{in}\mathrm{x}⩾\mathrm{a}\Rightarrow \underset{x\to {\mathrm{a}}^{+}}{\lim }\lfloor \mathrm{g}\left(\mathrm{x}\right)\rfloor =\lfloor \mathrm{g}\left(\mathrm{a}\right)\rfloor$$$$prove:\underset{x\to {\mathrm{a}}^{+}}{\lim g}\left(\mathrm{x}\right)=\mathrm{g}\left(\mathrm{a}\right)\Rightarrow \mathrm{\forall }\epsilon >0\mathrm{\exists }\sigma >0,$$$$0<\mathrm{x}-\mathrm{a}<\sigma \Rightarrow \mid \mathrm{g}\left(\mathrm{x}\right)-\mathrm{g}\left(\mathrm{a}\right)\mid <\epsilon \Rightarrow$$$$0<\mathrm{g}\left(\mathrm{x}\right)-\mathrm{g}\left(\mathrm{a}\right)<\epsilon \Rightarrow$$$$0⩽\lfloor \mathrm{g}\left(\mathrm{x}\right)\rfloor -\lfloor \mathrm{g}\left(\mathrm{a}\right)\rfloor ⩽\mathrm{g}\left(\mathrm{x}\right)-\mathrm{g}\left(\mathrm{a}\right)<\epsilon \Rightarrow$$$$\mid \lfloor \mathrm{g}\left(\mathrm{x}\right)\rfloor -\lfloor \mathrm{g}\left(\mathrm{a}\right)\rfloor \mid <\epsilon \Rightarrow \underset{x\to {\mathrm{a}}^{+}}{\lim }\lfloor \mathrm{g}\left(\mathrm{x}\right)\rfloor =\lfloor \mathrm{g}\left(\mathrm{a}\right)\rfloor$$$$,$$$$h\left(x\right)=nf\left(x\right)iscontinuousandincrement$$$$functioninx⩾c=0.5(3+\sqrt{5})\Rightarrow$$$$\underset{x\to c^{+}}{\lim }\lfloor h\left(x\right)\rfloor =\lfloor h\left(c\right)\rfloor =\lfloor \frac{n}{2}\rfloor =3$$$$\Rightarrow n=6or7$$

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