I am considering the uniform convergence of $(\tan(x))^n$ in the interval$[0,π/4)$. This function sequence is pointwise convergent everywhere in the given interval and converges to $0$.
Is this function uniform convergent in the given interval? In my opinion, it is uniform convergent because the supremum of $|f_n(x)-f(x)|$ as $n$ tends to $\infty$ is $0$. Then by the $M_n$ test, we can say that the given function sequence is uniformly convergent.
Had the interval been $[0,π/4]$, the function sequence would have pointwise converged to a discontinuous function and hence been not uniform convergent.
Uniform convergence of $\tan(x)^n$
real-analysisuniform-convergence
Related Solutions
For your third question, JohnD's answer in the page you are quoting, contains a trick that's often used in Analysis courses. That is, try to find:
$$\sup_{x\in[0,1]}|f_n(x)-f(x)|$$
The supmemum occurs at $x$ such that:
$$\frac{df_n(x)}{dx}=0$$
Solving the above for $x$, you get:
$$x=\pm\frac{1}{n}$$
$\frac{1}{n}\in[0,1]$, so substitute back to the function to get:
$$\sup_{x\in[0,1]}|f_n(x)|=f_n\left(\frac{1}{n}\right)=\frac{1}{2}$$
And this is fixed and does not vanish, therefore convergence in $[0,1]$ is not uniform.
Addendum for comment:
I am adding a graphic, so you can see what's happening as a response to your second question.
This is the graph of $f_1(x)$, $f_2(x)$,..., $f_5(x)$, from right to left. Note that the supremum is given by $\left(\frac{1}{n},f_n\left(\frac{1}{n}\right)\right)=\left(\frac{1}{n},\frac{1}{2}\right)$ and is moved to the left on each iteration, but always stays at 1/2.
For pointwise convergence, you fix $x\in [0,1] $ and you compute $\lim_ {n\to+\infty}f(x) $. There will be pointwise convergence in the set containing $x $ for which the limit exists $(\in\mathbb R) $. in your example,
for $x=0$, the limit is zero. for $x=1$, it is zero. for $0 <x <1$, write $(1-x)^n=e^{n\ln (1-x)} $ and you will find zero since exponential is faster than polynomial.
thus all the limits are zero in $[0,1] $ . $(f_n) $ converges (pointwise) to function $0$ at $[0,1] $
For uniform convergence
find the maximum of $|f_n (x)-0|=f_n (x)$ at $[0,1] $.
$$f'_n (x)=n^2 (1-x)^{n-1}(1-x-nx) $$
it is attained at $$x_n=\frac {1}{n+1} $$ $$f (x_n)=n (\frac {n}{n+1})^{n+1}$$
$$\lim_{n\to+\infty}f_n (x_n)=+\infty$$
the convergence is not uniform at $[0,1] $.
Best Answer
It is not uniformly convergent. Let $x_n=\arctan (1-\frac 1 n)$. Then $0<x_n <\frac {\pi} 4$ and $(\tan x_n)^{n}=(1-\frac 1 n )^{n} \to\frac 1 e$ so $\sup_x (tanx )^{n}$ does not tend to $0$.