Prove that if $a_n$ is a nonnegative sequance and: $\lim_{n\to \infty} a_n=a$ then $\lim_{n\to \infty} \sqrt[5]{a_n}=\sqrt[5]{a}$
I tried to do this using the definition of the limit of a sequence, but I am not entirely sure whether my reasoning is logically correct (btw. I don't know how to make this left/right arrow):
$ \lim_{n\to \infty} a_n=a \iff \forall\epsilon>0\ \exists n_0\in N \ \forall n>n_0 \ \ |a_n-a|< \epsilon$
$ \lim_{n\to \infty} a_n=a \iff \forall\epsilon>0\ \exists n_0\in N \ \forall n>n_0 \ \ |(\sqrt[5]{a_n}-\sqrt[5]{a})(\sqrt[5]{a_n^4}+\sqrt[5]{a_n^3a}+\sqrt[5]{a_n^2a^2}+\sqrt[5]{a_na^3}+\sqrt[5]{a^4})|< \epsilon$
$ \lim_{n\to \infty} a_n=a \iff \forall\epsilon>0\ \exists n_0\in N \ \forall n>n_0 \ \ |(\sqrt[5]{a_n}-\sqrt[5]{a})|< \frac{\epsilon}{\sqrt[5]{a_n^4}+\sqrt[5]{a_n^3a}+\sqrt[5]{a_n^2a^2}+\sqrt[5]{a_na^3}+\sqrt[5]{a^4}}$
$\lim_{n\to \infty} \sqrt[5]{a_n}=\sqrt[5]{a} \iff\forall\epsilon_1>0\ \exists n_1\in N \ \forall n>n_1 \ \ |\sqrt[5]{a_n}-\sqrt[5]{a}|< \epsilon_1 $
Where $\epsilon_1$ is what we received from the previous line