Showing $(a^{k_n})\to a^k$ whenever $(k_n)\to k$.

Question

If the sequence $(k_n)\to k$, is it possible to show $(a^{k_n})\to a^k$ without having to involve the theory of limits of functions.

i.e., to show that $\forall\varepsilon > 0, \exists K\in\mathbb{N}$ s.t $\forall n\ge K$ we have $\mid a^{k_n}-a^k\mid < \varepsilon.$

Rudin uses this result in one of the examples in a chapter which comes before concept of limits of functions.

Answer 1

Provided that $$ \left|a^{k_n}-a^k\right|=\left|a^k\left(a^{\left(k_n-k\right)}-1\right)\right|=\left|a^k\right|\left|a^{\left(k_n-k\right)}-1\right|, $$ you may then apply your $\epsilon$-$\delta$ argument with respect to $\left|k_n-k\right|$, as per your value of $a$ of course.