The question is as follows,
Given $b>1$ and $c \in \mathbb{R}$, find a sequence of step functions $f_n$ which converges uniformly to $f(x) = x^c$ on $[1, b]$. Use the partition $1 = b^{0/n} < b^{1/n} < \cdots < b^{n/n} = b$.
I have chosen my sequence $(f_n)_{n \in \mathbb{N}}$ as $$ f_n(x) = \begin{cases} \left(b^{k/n}\right)^c &: x \in [b^{k/n}, b^{(k+1)/n}) ~~ \text{(for $k = 0, \ldots, n-1$)} \\ b^c &: x = b \\ \end{cases} $$
Then I need to show $$(\forall \epsilon>0)(\exists N \in \mathbb{N})(\forall n \geq N)(\forall x \in [1, b]): |f_n(x)-f(x)|<\epsilon$$
In the case $x=b$ we simply have $$ |f_n(x)-f(x)|=|b^c-b^c|=0<\epsilon $$
But in the case $x \in [b^{k/n}, b^{(k+1)/n})$ we have $$ |f_n(x) - f(x)| = |b^{ck/n} - x^c| \leq |b^{ck/n}-b^{c(k+1)/n}| $$
Then how would I choose the lower bound on $n$ such that this is always less then $\epsilon$?