Let $f_n(x) = n\log(\frac{nx+1}{nx-1})$. What is the pointwise limit of $f_n$ for $x \in[1,\infty]$? Is it uniformly convergent?
$$\lim_{n\longrightarrow \infty}n\log(\frac{nx+1}{nx-1})=\lim_{n\longrightarrow \infty}\frac{\log\left(\frac{x+\frac{1}{n}}{x-\frac{1}{n}}\right)}{\frac{1}{n}}\stackrel{H}{=}\lim_{n\longrightarrow \infty}\frac{\frac{-2x}{n^2}\frac{1}{(x-\frac{1}{n})(x+\frac{1}{n})}}{\frac{-1}{n^2}}=\lim_{n\longrightarrow \infty}\frac{2x}{(x-\frac{1}{n})(x+\frac{1}{n})}=\frac{2}{x}$$ Is that correct? And how to check uniform covergence? I would really like not to search for $\sup$ of $|n\log(\frac{nx+1}{nx-1}) - \frac{2}{x}|$...