If `f(x) = lim_(n->oo) tan^(-1) (4n^2(1-cos(x/n))) `and g(x) = `lim_(n->oo) n^2/2 ln cos(2x/n)` then `lim_(x->0) (e^(-2g(x)) -e^(f(x)))/(x^6)` equals

Let f (x) =lim _(nto oo) tan ^(-1) (4n ^(2) (1- cos ""(pi)/(n )))and g (x) = lim _(n to oo) (n^(2))/(2) ln cos ((2x)/(n )) then lim _(xto0) (e ^(-2g (x))-ef (x))/(x ^(6)) equals.

lim_(n->oo)sin(x/2^n)/(x/2^n)

If f(x)=lim_(n rarr oo)(x^(2n)-1)/(x^(2n)+1) then range of f(x) is

lim_(x->oo)(1-x+x.e^(1/n))^n

let f(x)=lim_(n rarr oo)(x^(2n)-1)/(x^(2n)+1)

Let f(x)=(lim)_(n->oo)(2x^(2n)sin1/x+x)/(1+x^(2n))\ then find :\ (lim)_(x->-oo)f(x)

If f(x)=lim_(m->oo) lim_(n->oo)cos^(2m) n!pix then the range of f(x) is

lim_(x->oo)(e^x((2^(x^n))^(1/(e^(x)))-(3^(x^n))^(1/(e^(x)))))/(x^n), n in N, is equal to

If f(x)=lim_(n rarr oo)e^(tan((1)/(n))log((1)/(n))) and int(f(x))/(((sin^(11)x*cos x))^((1)/(3)))=g(x) then

f(n)=lim_(x->0){(1+sin(x/2))(1+sin(x/2^2)).......(1+sin(x/2^n))}^(1/x) then find lim_(n->oo)f(n)