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

The area of the triangle having vertices (-2,1), (2,1) and (lim_(mtoinftyntoinfty)cos^(2m)(n!pix) , x is rational lim_(mtoinftyntoinfty)cos^(2m)(n!pix) , where x is irrational) is

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)

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

lim_(n->oo)2^(n-1)sin(a/2^n)

If x is a real number in [0,1] . Find the value oif lim_(mtooo)lim_(ntooo)[1+cos^(2m)(n!pix)]

If f(x)=lim_(m rarr oo)lim_(n rarr oo)cos^(2m)n!pi x then the range of f(x) is

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

Let x be an irrational,then lim_(m rarr oo)lim_(n rarr oo){cos(n!pi x)}^(2m) equals