Question

If `f(x)=lim_(n->oo)1/(1+nsin^2pix)` , find the value of `f(x)` for all real values of `x`?

Answer 1

`f(x)=1` when `x` is an integer and `f(x)=0` otherwise.

Explanation:

If `x` is an integer (`x=0, pm 1, pm 2, pm 3,...`), then `pi x` is an integer multiple of `pi` so `sin(pi x)=0`. In this case, `1/(1+n sin^{2}(pi x))=1/(1+0)=1` for all `n`.

If `x` is not an integer, then `sin(pi x) !=0` so `n sin^{2}(pi x)->infty` as `n->infty`. In this case, `1/(1+n sin^{2}(pi x))->0` as `n->infty`.

The answer above follows.