Let x be an irrational, then `lim_(m->oo) lim_(n->oo) {cos(n!pix)}^(2m)` equals

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

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

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

lim_(m to oo)("cos"(x)/(m))^(m) is equal to

lim_(nto oo) (2^n+5^n)^(1//n) is equal to

lim_(nto oo) (2^n+5^n)^(1//n) is equal to

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

lim_(nrarr oo)(1+sin.(1)/(n))^n equals

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: