`lim_(x->oo)(1-sqrt(3)tanx)/(pi-6x)`

lim_(x->oo) (x-sqrt(x^2-1))

If (x)=(1-sqrt3tanx)/(pi-6x),"for"x nepi/6 is continous at x=pi/6, "find " f ((pi)/(6)).

lim_(x->oo)(sqrt(x+sqrt(x))-sqrt(x)) equals

lim_(x->(pi)/6)(3sinx-sqrt(3)cosx)/(6x-pi)

lim_(x->(pi/2)) (1-sinx)tanx =

lim_(x->(pi/2)) (1-sinx)tanx =

The value of lim_(xto0^(+))(-1+sqrt((tanx-sinx)+sqrt((tanx-sinx)+sqrt((tanx-sinx)+…oo))))/(-1+sqrt(x^(3)+sqrt(x^(3)+sqrt(x^(3)+…oo)))) is

lim_(x->oo)x^(3/2)(sqrt(x^3+1)-sqrt(x^3-1))