Question:
Prove that the sequence of functions $f_n(x) = x^n$ converges uniformly to zero on any interval of the form $[0,\mu]$ if $\mu < 1$.
My Work:
Recalling the definition of uniform convergence, we say that $f_n(x) \to 0$ uniformly if $\forall\epsilon>0 \ \exists N \ \forall n \ge N \ \forall x \in \mathrm{Dom}\,(f) \ \Big[\left|f_n(x) - 0 \right|\le \epsilon \Big] $
Since $x \in [0,\mu]$, we have $x \le \mu \Rightarrow \left| f_n(x) -f(x) \right| \le \mu^N$. We want $\mu^N \le \epsilon$ for some large $N$, so taking the natural logarithm of the inequality, which we can do because the natural log is monotone increasing for $x >0$, we have $ \mu^N \le \epsilon \Rightarrow N \ln \mu \le \ln \epsilon $ Since $\mu, \epsilon < 0$ for $\epsilon$ sufficiently small, $\ln \mu, \ln \epsilon < 0$. So the inequality becomes $ N \ge {\ln \epsilon \over \ln \mu}$
This makes sense to write because ${\ln \epsilon \over \ln \mu}$ is a positive number, so we can always take $N \ge {\ln \epsilon \over \ln \mu}$.
My Question: How do I incorporate this $N$ into an appropriately worded proof? I know it would go along the lines of "Taking $N \ge {\ln \epsilon \over \ln \mu}$, we have $ \mu^N - 0 \le \mu^{{\ln \epsilon \over \ln \mu}} $ By virtue of my previous calculation, I know this should be less than $\epsilon$, but I am not sure how to perform the necessary algebra. Its a small, but crucial detail.
Edit:
Thanks to Nate Eldrege, who pointed out that $a^b = \exp(b \ln a)$. So we have $ \mu^{{\ln \epsilon \over \ln \mu}} = \mu ^{\ln(\epsilon - \mu)} = e^{\ln(\epsilon - \mu)}e^{\ln \mu} = \epsilon $ So as required for any $n \ge N$, where the value of $N$ is given above, we have $\mu^N - 0 \le \epsilon$.