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

Let f(x)=lim_(m->oo){lim_(n->oo)cos^(2m)(n !pix)}, where x in Rdot Then the range of f(x)

Let f(x)=lim_(m->oo){lim_(n->oo)cos^(2m)(n !pix)}, where x in Rdot Then prove that f(x)={1, if x is rational and 0, if x is irrational

Let f(x)=lim_(m->oo){lim_(n->oo)cos^(2m)(n !pix)}, where x in Rdot Then prove that f(x)={1, if x is rational and 0, if x is irrational

Let f(x)= lim_(n->oo)(sinx)^(2n)

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

If x =u is a point of discontinuity of f(x)= lim_(n to oo) cos^(2n)x , then the value of cos u is

The area of the triangle having vertices (-2,1),(2,1) and lim_(m rarr oo)lim_(n rarr oo)cos^(2m)(n!pi x);x is rational,lim_(m rarr oo)lim_(n rarr oo)cos^(2m)(n!pi x); where x is irrational) is:

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

Let f(x) be defined as f(x)=max{a,b,c} where a=lim_(n rarr oo)lim_(alpha rarr1^(+))(alpha^(n)|sin x|+|cos x| alpha^(-n))/(alpha^(n)+alpha^(-n)) , b=lim_(n rarr oo)lim_(alpha rarr1^(-))(alpha^(n)|sin x|+|cos x| alpha^(-n))/(alpha^(n)+alpha^(-n)) , c=lim_(n rarr oo)(pi)/(4n)[1+cos(pi)/(2n)+...+cos((n-1)pi)/(2n)] then The range of f(x) is

Let f(x)=lim_( n to oo)(cosx)/(1+(tan^(-1)x)^(n)) . Then the value of int_(o)^(oo)f(x)dx is equal to