`("lim")_(xvecoo)(x(log"x")^3)/(1+x+x^2)e q u a l` 0 (b) `-1` (c) 1 (d) none of these

Options

A. `underset(xto0^(+))f(x)=1`
B. `underset(xto0^(-))limf(x)=cot1`
C. `cot^(-1)(underset(xto0^(-))f(X))^(2)=1`
D. `tan^(-1)(underset(xto0^(+))f(x))=(pi)/(4)`

Answer

A

Explanation

`underset(xtooo)lim((log)^(3)+x.3(logx)^(2)xx(1)/(x))/(1+2x)" "`(Using L'Hospital's rule)
`=underset(xtooo)lim(3(logx)^(2)xx(1)/(x)+6(logx)xx(1)/(x))/(2)`
`=underset(xtooo)lim(3(log)^(2)+6logx)/(2x)`
`=underset(xtooo)lim(6logxxx(1)/(x)+(6)/(x))/(2)" "`(Using L'Hospital's rule)
`=underset(xtooo)lim(6logx+6)/(2x)`
`=underset(xtooo)lim(6((1)/(x))+0)/(2)" "`(Using L'Hospital's rule)
`=((6)/(oo))/(2)=0`