For `x in (0,1)` Prove that `x-(x^(2))/(3) lt tan^(-1) x lt x -(x^(2))/(6)` hence or otherwise find `lim_( x to0) [(tan^(-1)x)/(x)]`

Explanation

`" Let " f(x) =x-(x^(3))/(3)- tan^(-1) x " " f(x) =1- x^(2) -(1)/(1+x^(2)) " "f(x) =-(x^(4))/(1+x^(2))`
`f(x) lt 0 " for " x in (0,1) " "rArr " " f(x) " is " M.D`
` rArr f(x) lt f(0) " "rArr" " x-(x^(3))/(3) -tan^(-1) x lt 0`
` rArr x-(x^(3))/(3) lt tan^(-1) x " ".........(1)`
`"Similarly g (x) "=x-(x^(3))/(6) -tan^(-1) x . g(x) =1-(x^(2))/(2)- (1)/(1+x^(2)) g (x) =(x^(2) (1-x^(2)))/(2(1+x^(2)))`
`g(x) gt 0" " "for " X in (0,1) rArr " " g(x) " is " M.I`
`rArr g(x) gt g(0) " " x-(x^(2))/(6) tan^(-1) x gt0`
` x=(x^(3))/(6) gt tan^(-1) x " ".........(ii)`
`"from (i) and (ii) , we get "" "x -(x^(3))/(3) lt tan^(-1) x lt x -(x^(3))/(6) " ""Hence Proved"`
`" Also "" "1-(x^(2))/(3) lt (tan^(-1))/(x) lt 1 -(x^(2))/(6) " for x gt 0`
Hence by sandwich theorem we can prove that `underset(h to0)("lim") .(tan^(-1)x)/(x)=1` but it must also be noted that as `x to 0`
value of `(tan^(-1)x)/(x) to1` from left hand side i.e., `(tan^(-1)x)/(x) lt 1`
`rArr underset(h to0)("lim") ,[(tan^(-1)x)/(x)] =0`
Note : In proving inequalities we must always check when does the equality takes place because the point of equality is very important in this method. Normally point of equality occur at end point of the interval or will be easily predicted by hit and trial.