Finding limits of indeterminate form using transformations I am trying to find the limit 
$\large\lim_{n \to \infty} (n^{\frac{1}{2n}})$
WolframAlpha says that I can transform as follows
$\large\lim_{n \to \infty} (n^{\frac{1}{2n}})=e^{\lim_{n \to\infty} \frac{ln(n)}{2n}}$
However, I do not understand where $n$ in $ln (n)$ comes from. Could anyone explain this?
 A: Note $n=e^{\log n}$.
It tries to use the continuity of exponential function. That is if $a_n\to a$, then $e^{a_n}\to e^a$, while here $a_n=\frac{ln(n)}{2n}$. Then the problem reduces to study the convergence of $a_n$
You can also find an elementary proof for $\lim_{n\to\infty}n^{\frac{1}{n}}=1$, and your sequence is just $\sqrt{n^{\frac{1}{n}}}$.
A: First note that
$$ n=e^{\ln |n|}, \forall n\in\mathbb{R} $$
In this case, since $n\to\infty$, then $n\gt 0$ and we can drop the absolute value bars. So we have
$$ n=e^{\ln n} $$
Which also implies that
$$ n^a=e^{\ln n^a} =e^{a\ln n}$$
Here are the steps
$$ \lim\limits_{n\to\infty} n^{\frac{1}{2n}}= \lim\limits_{n\to\infty} e^{\ln n^{\frac{1}{2n}}} = \lim\limits_{n\to\infty} e^{\frac{1}{2n}\ln n} = \lim\limits_{n\to\infty} e^{\frac{\ln n}{2n}}= e^{ \left(\lim\limits_{n\to\infty} \frac{\ln n}{2n}\right)}= e^{ \left(\lim\limits_{n\to\infty} \frac{\frac{d}{dn}[\ln n]}{\frac{d}{dn}[2n]}\right)}= e^{ \left(\lim\limits_{n\to\infty} \frac{\left(\frac{1}{n}\right)}{2}\right)}= e^{ \left(\lim\limits_{n\to\infty} \frac{1}{2n}\right)} =e^0=1$$
