Basic Limits Proof (Calculus )

Question

Show $\lim_{n \to \infty} (\alpha^{1/n})=1$

Proof in book: $\alpha^{1/n} = e^{1/n \ln{\alpha}}$. But $\ln{\alpha}/n \to 0 \implies \lim_{n \to \infty} (e^0) = 1$ (by continuity of function $f(x)= e^x$)

My Question: The exponential function $(\alpha^{1/n})$ is also continuous. Can't we just say that since $1/n \to 0$ then it's equivalent to $\lim_{n \to \infty} (\alpha^0) =1$ (by continuity of function $\alpha^x$)?

Answer 1

Yes. That's more or less what's done in the proof offered, just using base $e$ and the exponential. If it can be taken for granted...

• $\alpha^x$ is continuous on the reals for $\alpha > 0$ (your offered proof implicitly seems to assume $\alpha > 0$ as well, by the way)
• $1/n \to 0$ as $n \to \infty$

... there wouldn't be an issue with claiming through continuity that $\alpha^{1/n} \to 1$ as $n \to \infty$. At least none that I'm immediately aware of.