Uniform convergence and local uniform convergence of the sequence $n(x^\frac{1}{n}-1)$

Question

Why is $n(x^\frac{1}{n}-1)$ not uniformly convergent in $(0, \infty)$? I know that the given sequence converges pointwise to ln($x$). Further, is it locally uniformly convergent in every $(\frac{1}{k}, k)$ where $k \in \mathbb N$?

Answer 1

Uniformity of convergence on $(\frac 1 k,k)$ can be proved easily from the fact that $x^{1/n}=e^{\frac 1 n \ln x}=1+\frac 1 n \ln x +O(\frac 1 {n^{2}})$ uniformly on $(\frac 1 k,k)$.

Non-uniformity of convergence on $(0,\infty)$: pointwise limit is $\ln x$. If the convergence is uniform then there exists $N$ such that $|n(x^{1/n} -1)-\ln x | <1$ for all $x$ for all $n >N$. Put $x=e^{n}$ to get $n(e-2) <1$ for al $n >N$ which is a contradiction.