Limit of $((1+z_n)^{1/z_n})^{\log n}$ for sequence $z_n\to 0$.

Question

Let $(z_n)$ be any sequence with $\lim_n z_n=0$.

Then, what is

$$ \lim_{n\to\infty}\left((1+z_n)^{1/z_n}\right)^{\log n}? $$

By the linked question, I know that $$ \lim_{n\to\infty}(1+z_n)^{1/z_n}=e. $$

Nut how to handle the outer exponent $\log n$?

Answer 1

Simply note that it is not an indeterminate form thus

$$\left((1+z_n)^{1/z_n}\right)^{\log n}\to e^{+\infty}=+\infty$$

As an alternative note that

$$\left((1+z_n)^{1/z_n}\right)^{\log n}=e^{\log n \cdot \log \left((1+z_n)^{1/z_n}\right)}\to e^{+\infty\cdot 1}=+\infty$$