Prove or disprove that sequence of functions is uniform convergent I would like to see if $$f_n(x) = n^{-x}x^n\cos (nx)$$ converges uniformly on $[0,1]$
The sequence converges pointwise to $f(x)=0$.
By using the crieria the $f_n$ converges if and only if $$M_n = sup_{x\in[0,1]} |f_n(x) - f(x)|$$ converges to $0$, I obtain an equation for $x$ that is not solvable analytically.
So I though of using the fact that $$f_n(x) <=  n^{-x} x^n \forall x\in [0,1]$$.
By obtaining the $M_n$ for this new sequence, I get that the supremum is obtained by $x_0 = \frac{n}{\log(n)}$. 
What is then left to show is that $M_n =  n^\frac{-(n)}{\log(n)} \left(\frac{n}{\log (n)}\right)^n $ tends to $0$ as $n$ tends to infinity. This however, I am not able to prove (and I am not even sure this is the case).
Can someone help/give me a hint with this? Help for the starting question or/and the subsequent question which I ended up with are most welcome.
Thank you! 

Hint:
$|f_n(x)| = |n^{-x} x^n \cos (nx)| \le n^{-x} x^n$.
Both the factors on the right-hand side are $\le 1$ in absolute value; one of them tends to zero uniformly on $[0,\frac{1}{2}]$, and the other on $[\frac{1}{2},1]$.
