Smallest real number; big-O of a^x.

Question

Is there a smallest real number a for which (x^2)(6^x) is big-O of a^x?

Again, I'm not understanding my class lecture notes, and any assistance on how to go about this question is greatly appreciated.

Answer 1

Definition: f(x) is in O(g(x)) iff there exists a k>0 and an x₀>0 so that for every x>x₀, |f(x)| ≤ k*g(x).

In other words, the ratio |f(x)|/g(x) gets upper bounded by some constant k.

Consider a=6. No matter what k we pick, eventually |x^2 * 6^x| > k*6^x for a large enough x.

|x^2 * 6^x| / 6^x = x^2 for x>0, and x^2 cannot be upper bounded by a constant.

Therefore (x^2)(6^x) is not big-O of 6^x.

Consider any arbitrary a>6.

The limit of |x^2 * 6^x| / a^x as x goes to infinity is 0. So there exists a sufficiently large value of x₀ so that for all x>x₀, |x^2 * 6^x| / a^x is upper bounded by some k>0, the limit.

So for all a<=6, we have (x^2)(6^x) is not big-O of a^x.

And for all a>6, we have (x^2)(6^x) is big-O of a^x.

However, there is no smallest real number greater than 6.