Proof of $\lim_{n \to \infty} a^{1/n} = 1$ and $\lim_{n \to \infty}b^{f(n)} = b^{\lim_{n \to \infty} f(n)}$ where $a>1$ As title says, I am not sure what would be the proof of $$\lim_{n \to \infty} a^{1/n} = 1$$ would be where $a>1$. Also, how do you prove that $$\lim_{n \to \infty}b^{f(n)} = b^{\lim_{n \to \infty} f(n)}$$ where $b>0$?

1) If you want to argue more formally. For $a>1$, we have $a^{1/n}>1$. Define $c_n = a^{1/n}-1>0$, then $a=(1+c_n)^n\ge1+nc_n$. Thus
$$0<c_n= a^{1/n}-1\le \frac{a-1}{n}\to0$$
Hence  $a^{1/n}-1\to 0$, i.e., $a^{1/n}\to 1$.
2) Try to prove that $b^x$ is continuous if you want I could give a sketch of the proof. 
