Does $f_n$ uniformly converge to $f$? $f_n :[0, 1]\rightarrow\mathbb{R} \qquad x \mapsto x^n - x^{n+1}$
The sequence converges pointwise to the zero function. It converges uniformly if
$$\sup_{x \in [0, 1]} \; \big| \, f_{n}(x) - f(x) \, \big|$$
tends to zero. But I am not sure if it does or how to prove.

Note that $f_n'(x)=nx^{n-1}-(n+1)x^n=x^{n-1}\bigl(n-(n+1)x\bigr)$. Therefore, the maximum of $f_n$ is $f_n\left(\frac n{n+1}\right)$. But$$f_n\left(\frac n{n+1}\right)=\left(\frac n{n+1}\right)^n\frac1{n+1}.$$Since$$\lim_{n\to\infty}\left(\frac n{n+1}\right)^n\frac1{n+1}=0,$$your sequence converges uniformly.
