Past exam of the mathematics course of the University of Cambridge / 2007 / ib / Paper 4 / 3H (Analysis II)

The sequence has uniform convergence to $f$ on $(0,1)$ precisely whenThe same $N$ must work throughout the interval; equivalently ${\sup }_{0<x<1}|f_n(x)-f(x)|\to 0$.

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 $n>1$,Consequently the uniform error cannot tend to zero. In fact $f_n(x)-\log x=\log (1+1/(nx))$ is unbounded as $x\downarrow 0$ for every fixed $n$. The pointwise limit is $\log x$, and convergence is not uniform.

For each fixed $x$, $x/n\to 0$, so the pointwise limit is $1$. Moreover $1-\cos u=2{\sin }^2(u/2)\le u^2/2$ givesThus the limit is $1$, with uniform convergence on $(0,1)$.