I'm having doubts about the pointwise limit of the following function: $f_n(x)=(\tan(x))^n$ with $x\in [0,\frac{\pi}4]$
I know that $f_n$ converges to $0$ when $x∈ [0,\frac{\pi}4[$. However, when $x=\frac{\pi}4$, $f_n$ converges to 1. So it seems that $f_n$ converges pointwise to
$$f(x)=\begin{cases}
0\mbox{ if $0\leq x<\pi/4$,}\\
1\mbox{ if $x=\pi/4$.}
\end{cases}$$
So because my $f(x)$ isn't continuous i'm not sure what the pointwise limit is.
I'm then going to need to calculate my $M_n$, which is the supremum of $|f_n(x)-f(x)|$, but because I'm not understanding the pointwise limit i don't know what to substitute $f(x)$ for or what the interval of $x$ related to the supremum is.
Can anyone help me?
Best Answer
The pointwise limit $f$ is correct. Since $f$ is not continuous, it follows that $(f_n)_n$ does not converge uniformly to $f$ in $[0,\pi/4]$ (actually neither in $[0,\pi/4)$). What about the uniform convergence in the interval $[a,b]$ with $0<a<b<\pi/4$?