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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.

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up vote 5 down vote accepted

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.

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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 $[a,\infty)$.

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