I have shown that $f_n\left(x\right)=n\left(\sqrt[n]{x}-1\right)$ converges pointwise to $f\left(x\right)=\ln x$ on $\left(0,\infty\right)$, and I am now trying to show that it converges uniformly on $\left[e^{-A},e^A\right]$.
Recall the definition of uniform convergence:
Let $f_n$ be a sequence of functions defined on a set $A\subseteq\mathbb{R}$. Then, $\left(f_n\right)$ converges uniformly on $A$ to a limit function $f$ defined on $A$ if, for every $\epsilon>0$, there exists an $N\in\mathbb{N}$ such that $\left|f_n\left(x\right)-f\left(x\right)\right|<\epsilon$ whenever $n\geq N$ and $x\in A$.
One thing I noticed I could use is that if $x\in\left[e^{-A},e^A\right]$ and $y=\ln x$, then $y\in\left[-A,A\right]$. However, after having tried that out, I have run of ideas as to how to begin to tackle this problem altogether. Could you guys give me a few pointers? Thanks in advance!
Edit 1: I have tried using the triangle inequality: $\left|n\left(\sqrt[n]{x}-1\right)-\ln x\right|\leq\left|n\left(\sqrt[n]{x}-1\right)\right|+\left|\ln x\right|.$And since $x\in\left[e^{-A},e^A\right]$: $\left|n\left(\sqrt[n]{x}-1\right)\right|+\left|\ln x\right|\leq\left|n\left(\sqrt[n]{x}-1\right)\right|+\left|A\right|=n\left(\sqrt[n]{x}-1\right)+A.$ Does this seem right? If so, all that I am left to do is show that for any $\epsilon>0$, there exists an $N\in\mathbb{N}$ such that the last expression above is less than $\epsilon$ whenever $n\geq N$ and $x\in \left[e^{-A},e^A\right]$. But I am stuck. What do you guys think?