Question

If f(x)=lim of 1/(1+nsin^2pix) as n approaches to infinity,find the value of f(x) for all real values of x?

Answer 1

`f(x) = \mathcal(I)(ZZ)` - a function that returns 1 when `x in ZZ`, 0 otherwise.

Explanation:

If `x in ZZ`, `sin^2(pi x)=0` and so

`f(x) = lim_{n to oo} 1/(1+n sin^2(pi x))=lim_{n to oo} 1/(1+n times 0)`
`qquad = lim_{n to oo} 1/(1+0) = 1`

If `x notin ZZ`, `sin^2(pi x) > 0` and so

`f(x) = lim_{n to oo} 1/(1+n sin^2(pi x))=lim_{n to oo} (1/n)/(1/n+sin^2(pi x) )`
`qquad = 0/sin^2(pi x) = 0`