I want to show that $f_n(x) = x^n \log(1+\frac1{x^n})$ converges uniformly to $0$ in the interval $(\epsilon,\delta), 0<\epsilon<\delta<1$ and to $1$ in the interval $(a,\infty), 1<a$.
So far, to prove convergence in the second interval I have considered $$\lim_{n \to \infty}\|f_n-1\|_\infty = \lim_{n \to \infty}\|x^n \log(1+\frac1{x^n})-1\|_\infty$$
As $\log(1+z)$ is an infinitesimal equivalent of $z$ when $z \to 0$, we can change one for another inside the \limit: $$=\lim_{n \to \infty}\|x^n \frac1{x^n}-1\|_\infty = 0$$
Regarding the first interval, I tried as well to evaluate $$\lim_{n \to \infty}\|f_n\|_\infty = \lim_{n \to \infty}\|x^n \log(1+\frac1{x^n})\|_\infty = \|0 \cdot \infty\|_\infty$$ L'Hôpital's rule came to me, but when applied: $$=\lim_{n \to \infty}\|\frac{(\log(1+\frac1{x^n}))'}{(\frac1{x^n})'}\|_\infty=\lim_{n \to \infty}\|\frac1{1+\frac1{x^n}}\|_\infty=\|\infty\|_\infty$$
May I get a little help?
Your evaluation of the limit after l'Hospital is wrong:
$$...\lim_{n\to \infty}\frac{x^n}{1+x^n}=\frac0{1+0}=0\;,\;\;\text{since}\;\;|x|<1$$