Prove that this sequence does NOT converge uniformly on [0,1] AND (0,1). I'm not familiar with LaTeX so I understand if this topic is not formatted correctly and thus must be removed.  The question at hand and my current work will be attached as an image.  Thanks to anyone that can help!

 A: The function
$$
   \phi_n(x) = \frac{x}{2^n - (2^n-1)x}
$$
has a pointwise limit $\phi(x)$ at every $x$. If it's going to converge to something uniformly, it's going to be to $\phi(x)$.
You should figure out what $\phi(x)$ is as a function of $x$. (Your work for (a) shows how to find $\phi(1)$, since you substitute $x=1$ in the first step.) 
For either part (a) or part (b), your approach can be used to show that $\phi_n$ does not uniformly converge to $\phi$. Except instead of using $1$, you should use the correct value of $\phi(x)$. If you show that the sequence does not converge uniformly on $(0,1)$, then it does not converge uniformly on $[0,1]$ either. The only difference between the two solutions is that the value
$$
   \sup\{|\phi_n(x) - \phi(x)| : x \in S\}
$$
allows $x$ to be $0$ or $1$ when $x\in[0,1]$, and does not allow it when $x\in(0,1)$. If you don't need $0$ or $1$ for computing the supremum, then the same argument works in both cases.
However, for (a) a shorter argument also exists, using the fact that $\phi$ is not continuous on $[0,1]$. And for (b), you can use (a) together with this result. So even though there is an argument that solves both cases simultaneously, there is a different argument that does not, which might be easier.
