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f-R-R-is-defined-by-f-x-1-1-if-x-Z-if-x-Z-Is-f-continuous-at-x-1-and-x-3-2-




Question Number 25825 by rita1608 last updated on 15/Dec/17
f:R→R is defined by   f(x)={1_(−1  if x∉Z)    if x∈Z  Is f continuous at x=1 and x=−(3/2) ∫?
$${f}:{R}\rightarrow{R}\:{is}\:{defined}\:{by}\: \\ $$$${f}\left({x}\right)=\left\{\underset{−\mathrm{1}\:\:{if}\:{x}\notin{Z}} {\mathrm{1}}\:\:\:\mathrm{if}\:\mathrm{x}\in{Z}\right. \\ $$$${Is}\:{f}\:{continuous}\:{at}\:{x}=\mathrm{1}\:{and}\:{x}=−\frac{\mathrm{3}}{\mathrm{2}}\:\int? \\ $$$$ \\ $$
Answered by prakash jain last updated on 15/Dec/17
lim_(x→1^+ ) f(x)=lim_(h→0^+ ) f(1+h)=−1       ∵ (1+h)∉Z as h→0  similary  lim_(x→1^− ) f(x)=−1  f(1)=1  LHL=RHL≠f(1) function  is not continuius at 1.  for 3/2  LHL=RHL=−1  f(3/2)=−1  f(x) is continous at 3/2
$$\underset{{x}\rightarrow\mathrm{1}^{+} } {\mathrm{lim}}{f}\left({x}\right)=\underset{{h}\rightarrow\mathrm{0}^{+} } {\mathrm{lim}}{f}\left(\mathrm{1}+{h}\right)=−\mathrm{1}\: \\ $$$$\:\:\:\:\because\:\left(\mathrm{1}+{h}\right)\notin\mathbb{Z}\:\mathrm{as}\:{h}\rightarrow\mathrm{0} \\ $$$$\mathrm{similary} \\ $$$$\underset{{x}\rightarrow\mathrm{1}^{−} } {\mathrm{lim}}{f}\left({x}\right)=−\mathrm{1} \\ $$$${f}\left(\mathrm{1}\right)=\mathrm{1} \\ $$$$\mathrm{LHL}=\mathrm{RHL}\neq{f}\left(\mathrm{1}\right)\:\mathrm{function} \\ $$$$\mathrm{is}\:\mathrm{not}\:\mathrm{continuius}\:\mathrm{at}\:\mathrm{1}. \\ $$$$\mathrm{for}\:\mathrm{3}/\mathrm{2} \\ $$$$\mathrm{LHL}=\mathrm{RHL}=−\mathrm{1} \\ $$$${f}\left(\mathrm{3}/\mathrm{2}\right)=−\mathrm{1} \\ $$$${f}\left({x}\right)\:\mathrm{is}\:\mathrm{continous}\:\mathrm{at}\:\mathrm{3}/\mathrm{2} \\ $$

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