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Question Number 42897 by Joel578 last updated on 04/Sep/18

Commented by Joel578 last updated on 04/Sep/18

$$\mathrm{For}\mathrm{question}\left(c\right),$$$$\underset{x\to 0}{\lim }G\left(x\right)=2\mathrm{or}\underset{x\to 0}{\lim }G\left(x\right)=0?$$

Answered by MJS last updated on 04/Sep/18

$$\mathrm{the}\mathrm{limit}\mathrm{is}1\mathrm{although}G\left(0\right)=0$$

Answered by tanmay.chaudhury50@gmail.com last updated on 04/Sep/18

$$a)\underset{x\to 0-}{\lim }G\left(x\right)=1$$$$b)\underset{x\to 0+}{\lim }G\left(x\right)=1$$$$now\underset{x\to 0-}{\lim }G\left(x\right)iscalledlefthandsidelimit$$$$\underset{x\to 0+}{\lim }G\left(x\right)iscalledrighthandsidelimit$$$$limitexistif\underset{x\to 0-}{\lim }G\left(x\right)=\underset{x\to 0+}{\lim }G\left(x\right)$$$$thatis\underset{x\to 0}{\lim }G\left(x\right)=\underset{x\to 0-}{\lim }G\left(x\right)=\underset{x\to 0+}{\lim }G\left(x\right)$$$$here\underset{x\to 0-}{\lim }G\left(x\right)=1=\underset{x\to 0+}{\lim }G\left(x\right)$$$$sothevalueof\underset{x\to 0}{\lim }G\left(x\right)=1$$$$soc)\underset{x\to 0}{\lim }G\left(x\right)=1$$$$butthefunctionisdiscontinueatx=0$$$$soG\left(0\right)cannotbefoundout$$$$wecannotfindthevalueofG\left(x\right)atx=0$$$$d)cannotbefindthevalueofG\left(x\right)atx=0$$$$thatisG\left(0\right)notdefined$$$$thecurveG\left(x\right)isdiscontinueatx=0$$$$furtherwecansayG\left(x\right)isdiscontinueatx=0$$$$wecannotfind\frac{dG\left(x\right)}{dx}atx=0$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$c)\underset{x\to 0}{\lim }G\left(x\right)=\underset{x\to 0-}{\mathrm{li}}\underset{x\to 0}{\mathrm{li}}$$

Commented by Joel578 last updated on 04/Sep/18

$$\mathrm{Okay},\mathrm{thank}\mathrm{you}\mathrm{for}\mathrm{clearing}\mathrm{my}\mathrm{doubt}$$

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