Solution
= Solution
For each fixed $x>0$, <continuity> of the <logarithm> gives the <pointwise convergence> $f_n(x)\to\log x$. But at $x_n=1/n$, which belongs to $(0,1)$ for $n>1$,
$$
f_n(x_n)-\log x_n=\log\frac{2/n}{1/n}=\log2.
$$
Consequently the uniform error cannot tend to zero. In fact $f_n(x)-\log x=\log(1+1/(nx))$ is unbounded as $x\downarrow0$ for every fixed $n$. \b[The <pointwise limit> is $\log x$, and convergence is not uniform.]