Since, `underset(xtooo)lim(a)/(2^(x))=0,` we have `underset(xtooo)lim2^(x-1)tan((a)/(2^(x)))" "`(`ooxx0` form) `=underset(xtooo)lim(a)/(2).tan((a)/(2^(x)))/(((a)/(2^(x))))" "`(0/0 from)`=a/2xx1=a/2`
Evaluate (lim_(x->oo))(x(log)_e{(sin(a+1/x))/(sina)}) , 0 lt a lt pi/2
lim_(xtooo) (1)/(x+1)tan((pix+1)/(2x+2)) is equal to
If int(x(x-1))/((x^(2)+1)(x+1)sqrt(x^(3)+x^(2)+x))dx =(1)/(2)log_(e)|(sqrt(f(x))-1)/(sqrt(f(x))+1)|-tan^(-1)sqrt(f(x))+C, then The value of lim_(x to oo) tan^(-1)sqrt(f(x)) is
