Question

Regarding Big O notation, if the limit from n to infinity of |f(n)|/|g(n) approaches 0, will f(n) still = O(g(n))?

Answer 1

what is Big O .. is it order?

Answer 2

ok .. I got it ..

see if you have f(x)/g(x) tending to 0 for x --> infinity it means that there is some m for which for all x > m (f(x)/g(x)) is a decreasing function. Now assuming that there is a M which is only slightly greater than m. Hence for all x > m we would have f(x)/g(x) < f(M)/g(M)

Now assume f(M)/g(M) = N (some number)

then for all x> m we have f(x)/g(x) < N. Hence f(n) = O(g(n))