Past exam of the mathematics course of the University of Cambridge / 2025 / ib / Paper 2 / 2G (Analysis and Topology)

Pointwise, $\tanh (nx)$ tends to $-1,0,1$ according as $x<0,x=0,x>0$. This limit is discontinuous, whereas a uniform limit of continuous functions is continuous. Convergence is therefore not uniform.

The limit is $\sin x$, andso convergence is uniform.

The pointwise limit is $f(x)=e^x/\cosh (ax)$ andThe final factor is bounded on $\mathbb{R}$ exactly when $|a|\ge 1$; when $|a|<1$ it diverges as $x\to +\infty$. Thus convergence is uniform exactly for $|a|\ge 1$.

The fundamental theorem of calculus givesThus $f_n(x)\to 4h^{\prime }(x)$. Uniform continuity of $h^{\prime }$ makes the difference from $4h^{\prime }(x)$ uniformly small over every interval of radius $1/n$, so convergence is uniform on all of $\mathbb{R}$.