I am given the sequence of functions $f_n(x)=x^n - x^{2n}$ on $[0,1]$. I must define a function $f(x)$ as a pointwise limit function on the indicated interval. If it is uniform, I must then find a sequence of real numbers $B_n \rightarrow 0$ such that $|f_n(x) - f(x)| \leq B_n$ for all $x$ in the interval. If not, I must find an $\epsilon > 0$ and a sequence $x_n$ such that $|f_n(x_n) - f(x_n)| \geq \epsilon$ for all $n$.
My attempted solution: I deduced that $f(x) = 0$ for $x \in [0,1]$. I looked to see if such a $B_n$ existed, so I took the derivative of $f_n(x)$ and found the maximum to occur at $x=(\frac{1}{2})^{\frac{1}{n}}$. Using this as $B_n$ we get that $B_n \rightarrow 1$ as $n \rightarrow \infty$. Therefore, $f_n(x)$ should not uniformly converge (I am not sure if this is a correct conclusion).
Assuming my previous conclusion is correct, I am having a hard time choosing an $\epsilon$ where I can show $|f_n(x_n) - f(x_n)| = \epsilon$ to show it is not uniformly convergent. Algebra is and has always been my weakest skill, so I am not sure if I am missing a clever choice, or if I did something wrong in my problem. Any help is appreciated, please let me know if I need to elaborate more on something. Thanks.
It's more simple than that. Since$$f_n\left(\sqrt[n]{\frac12}\right)=\frac14,$$just use $\varepsilon=\frac14$ to prove that the convergence is not uniform.