If $f_n(x) = n^{-x}x^n\cos(nx)$ defines the sequence of functions $\{f_n\}_{n \in \mathbb{N}}$. Does the sequence of function converge uniformly on $[0,1]$?
Should I just do the pointwise convergence, and then determine the uniform convergence by evaluating $M_n:= \sup \{ |f_n(x) - f(x)| \}$?
If taking the path of pointwise convergence, I get stuck with $f_n(1)$ as it leaves me with $\cos(n)$ on the RHS.
Thank you.
My work
$f_n(0) = 0$, $f_n(1) = 0$, and $\lim_{n \to \infty} f_n(x) = 0$ (using L'Hospital's Rule).
Then, $M_n := \sup\{|f_n(x) - f(x)|: x \in [0,1]\} = \max \{\frac{x^n\cos(nx)}{n^x}; 0 \le x \le 1\}$
$f_n'(x) = 0\Rightarrow x = 0$; so $M_n = f_n(0) = 0 \to 0$ as $n \to \infty$, obviously.
Therefore as $M_n \to 0$, as $n \to \infty$ we can conclude that $f_n$ is uniformly convergent on $[0, 1]$.
Actually the problem is not $\cos(nx)$ which is uniformly bounded by $1$, but the behaviour of the product $x^n\cdot n^{-x}$ near $0^+$ and $1^-$.
Hint. Note that if $x\in[0,1/2]$ then $$|f_n(x)|=\frac{x^n|\cos(nx)|}{n^x}\leq x^n\leq \frac{1}{2^n}.$$ On the other hand if $x\in[1/2,1]$ then $$|f_n(x)|=\frac{x^n|\cos(nx)|}{n^x}\leq \frac{1}{n^x}\leq \frac{1}{\sqrt{n}}.$$