lim n-->infinity 1^n

Question

Is lim n −>infinity 1^n simply 1? I know 1^infinity is indeterminate form, but I read somewhere that you only have to apply log differentiation if it is a function raised to another function. If this is wrong, how would I go about solving this?

Answer 1

There is nothing to solve: 1ⁿ = 1 for all values of n and the limit cannot be anything else; there are infinitely many values of 1ⁿ near 1 for any ε>0.