Given the sequence [dⁿ/dxⁿ b^x], where b is a real number...?

Question

Given the sequence [dⁿ/dxⁿ b^x], where b is a real number, find the limit of the sequence as n approaches infinity.

^_^

Answer 1

The first derivative of b^x is ln b b^x.... The second is ln b ln b b^x or (ln b)²b^x ... The third is (ln b)³b^x... The nth is (ln b)^n b^x....
If b < e, then the limit is 0.
If b = e, then the limit is e^x.
If b > e, then the limit is ∞.

^^!!

Answer 2

Assume b is a positive real number. The sequence will be {b^x, b^x(ln b), b^x(ln b)(ln b), ....}. As n approaches infinity, the sequence will tend to either 0, b^x, or infinity. If b < e, the sequence will approach 0, regardless of b and independent of x. If b = e, each term of the sequence will be identically b^(x). If b > e, the sequence will approach positive infinity, regardless of b and independent of x.