How $\lim_{n\to\infty}\frac{|x|}{(1+x^{2n})^{\frac{1}{n}}}=\frac{1}{|x|}\lim_{n\to\infty}\frac{1}{(1+\frac{1}{2^n})^{\frac{1}{n}}}$?

Question

Show that $\sum_\limits{n=1}^{\infty}\frac{x^n}{1+x^{2n}}$, for $x\in\mathbb{R}$.

Resolution: For $|x|>1$. Applying the root rest $\lim_{n\to\infty}\frac{|x|}{(1+x^{2n})^{\frac{1}{n}}}=\frac{1}{|x|}\lim_{n\to\infty}\frac{1}{(1+\frac{1}{2^n})^{\frac{1}{n}}}=\frac{1}{|x|}$

I tried to divide both in the numerator and in the denominator by $\frac{1}{|x|}$ but it does not deliver the result.

Question:

How does the author go from $\lim_{n\to\infty}\frac{|x|}{(1+x^{2n})^{\frac{1}{n}}}$to$\frac{1}{|x|}\lim_{n\to\infty}\frac{1}{(1+\frac{1}{2^n})^{\frac{1}{n}}}$ ?

Answer 1

It seems there is a typo with $2^n$ instead of $x^{2n}$, indeed note that

$$\lim_{n\to\infty}\frac{|x|}{(1+x^{2n})^{\frac{1}{n}}}=\frac{1}{|x|}\lim_{n\to\infty}\frac{x^2}{(1+x^{2n})^{\frac{1}{n}}}=\frac{1}{|x|}\lim_{n\to\infty}\frac{1}{(1+\frac{1}{x^{2n}})^{\frac{1}{n}}}=\frac{1}{|x|}$$