Does $f_n(x)=n(x^{\frac{1}{n}}-1)$ converge uniformly in $(0,\infty)$ and in $[0,10]$? I'd like your help with the deciding whether the following function uniformly converges in two intervals.
$$f_n(x)=n \left(x^{\frac{1}{n}}-1 \right) .$$
In $(0,10]$, with l'Hôpital, I showed that for every $x$ the limit of the function is $\ln x$, so with a help from Dini's theorem, I am able that the function is monotone and the limit is continuous so the function uniformly converges.
I'm not sure what should I do with the other interval. I tried to evaluate the limit of $|f_n(x)-f(x)|$, I didn't come to any smart conclusion. 
 A: The sequence does converge uniformly to $\ln x$ on any compact subset of $(0,\infty)$, but the convergence is not uniform in $(0,a\,]$ for any $a>0$. If the convergence were uniform, there would exist an $n_0$ such that for all $n\ge n_0$
$$
|f_n(x)-\ln x|\le1\quad\forall x\in(0,a\,].
$$
But
$$
|f_n(2^{-n})-\ln(2^{-n})|=\Bigl(\ln2-\frac12\Bigr)n
$$
is not bounded.
Observe that $(0,a\,]$ is not compact and Dini's theorem does not apply.
A: As pointed out by Julian, the sequence does not converge uniformly on $(0,a]$ for any $a>0$; but it does converge uniformly on any compact subset of $\Bbb R$.
For sets of the form $[a,\infty)$:
Using L'Hopital, you can show the pointwise limit is $f(x)=\ln x$ on $[a,\infty)$.
Now take $x_n=2^n$. Then
$$
|f_n(x_n)-\ln(x_n)|= |n(2-1) -  n\ln 2|=n(1-\ln2)\quad
\buildrel{n\rightarrow\infty}\over\longrightarrow\quad\infty.
$$
Thus, there is no $N$ so that $|f_n(x)-\ln x|<1$ for all $n\ge N$ and all $x\in [a,\infty)$.
Thus, the convergence is not uniform on $[a,\infty)$.
