Find the value of 1/x for the given values of x
(i) `x gt 3`
(ii) `x lt -2`
(iii) `x in (-1 , 3 ) - {0}`

Explanation

(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)`