Limit of $f_n(x) = n\log(\frac{nx+1}{nx-1})$ and its uniform covergence 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 $|n\log(\frac{nx+1}{nx-1}) - \frac{2}{x}|$...

Looks right to me. The supremum check is pretty easy here. Note that,
\begin{align}
g(x) &:= \frac 2x - n\log(\frac{nx+1}{nx-1}) \\
&= \frac 2x - n\log(nx+1) + n\log(nx-1) \\
\implies g'(x) &= -\frac 2{x^2} - \frac {n^2}{nx+1} + \frac {n^2}{nx-1} \\
&= -\frac 2{x^2} + \frac {2n^2}{(nx+1)(nx-1)} \\
&= -\frac 2{x^2} + \frac {2n^2}{n^2x^2-1} \\
&= -\frac 2{x^2} + \frac {2}{x^2-\frac 1{n^2}} \\
&\geq 0
\end{align}
So $g(x)$ is monotonic and approaches it's supremum at an endpoint. It's pretty clear that $\lim_{x\to\infty}g(x)=0$, so $\sup_{x\in[1,\infty)}|g(x)| = |g(1)| = |2-n\log(\frac{n+1}{n-1})| \underset{n\to\infty}{\longrightarrow} 0$
