Find the values of 1/x for the given values of x. `(i) x gt 3 (ii) x lt -2 (iii) x in (-1, 3) -{0}`
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(i) We have `3 lt x lt oo` `rArr 1/3 gt 1/x gt 1/(rarr oo) ( rarr oo " mean tends to infinity ")` `rArr 0 lt 1/xlt 1/3 ` (ii) We have `- oo lt xlt -2` `rArr (1)/(rarr - oo) gt 1/x gt 1/(-2)` `rArr 0 gt1/xgt-1/2 rArr -1/2 lt 1/x lt 0 ` `x in (-1 , 3)-{0}` , `rArr x in (-1 ,0) cup(0,3)` For x in (-1,0), we have `1/1 gt 1/xgt1/(rarr0^-)` (here `rarr 0^+)` means the value of x approches to 0 from left hand side or negative side `rArr -1 gt 1/xgt-oo` `rArr -oolt1/xlt-1` For `x in (0,3)` we have `1/(rarr 0^+)gt1/x gt 1/3` (here `rarr 0^+)` means the value of x approches to 0 from left hand side or negative side ) `rArr oo gt 1/xgt 1/3` `rArr 1/3lt1/xlt oo` From (1) and (2) `1/x in (-oo,-1)cup(1/3,oo)`
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