Finding the $n$ to show $f_{n}(x) = (\frac{x}{2})^{n} + (\frac{1}{x})^{n}$ converges pointwise and uniformly Finding the $n$ to show $f_{n}(x) = (\frac{x}{2})^{n} + (\frac{1}{x})^{n}$ converges pointwise within the interval $[1,2]$. 
So there are two cases:
i) when $1 < x < 2$, here $f(x) = 0$
ii) when $x = 1$ or $x = 2$
I'm working on case i) first and my issue is in trying to find the appropriate $n$. So I have arrived at the following:
$$\Bigg|\Bigg(\frac{x}{2}\Bigg)^{n} + \Bigg(\frac{1}{x}\Bigg)^{n} - 
0 \Bigg|= \Bigg(\frac{x}{2}\Bigg)^{n} + \Bigg(\frac{1}{x}\Bigg)^{n} < \frac{x^{2n} + 2}{2^{n}x^{n}} < \epsilon$$
After some manipulation I find myself at:
$$2 < x^{n}(2\epsilon - x^{n})$$
ii) To show that  $f_{n}$ converges uniformly, we have to restrict ourselves to the interval $(1,2)$. Doing this we are presented with the following expression: 
$$\|f_{n}(x) - f(x) \|_{\infty} = \sup_{x \in (1,2)}\Bigg|\Bigg(\frac{x}{2}\Bigg)^{n} + \Bigg(\frac{1}{x}\Bigg)^{n} - 
0 \Bigg| $$
I know the idea I want to use in my argument:  "Since $x \in (1,2)$ every value of $x$ put into that expression will create a set of fractions that are both less than 1. So taking the limit of these both they would both go to 0."
Now how would I write this out with the correct notation?
 A: Given $\varepsilon>0$, you can choose $n_1$ such that for $n\ge n_1$
$$\left(\frac{x}{2}\right)^{n}<\frac\varepsilon2,$$
for instance, any $n_1$ such that
$$n_1 >\frac{\log\left(\frac\varepsilon2\right)}{\log\left(\frac{x}{2}\right)}$$
(we used that $\frac x2<1$ so $\log\left(\frac x2\right)<0$).
In the same way, using that $\frac1x<1$ you can find $n_2$ such that if $n\ge n_2$ then
$$\left(\frac1x\right)^n<\frac\varepsilon2.$$
So if you choose $n_0=\max\{n_1,n_2\}$, for $n\ge n_0$ (which implies both $n\ge n_1$ and $n\ge n_2$), you have
$$\left|\left(\frac{x}{2}\right)^{n} + \left(\frac{1}{x}\right)^{n} - 
0 \right|= \left(\frac{x}{2}\right)^{n} + \left(\frac{1}{x}\right)^{n} < \frac\varepsilon2+\frac\varepsilon2=\varepsilon.$$
When you have to bound a sum of two or more terms, all of which you know tend to $0$, it is a common trick to choose an $n$ for each term dividing $\varepsilon$ for the number of terms and then take the bigger $n$.

NOTE: If you wanted to prove uniform convergence, you should find an $n_0$ depending on $\varepsilon$ but not depending on $x$. The $n_0$ we gave here clearly does not satisfy that condition (because is the maximum between $n_1$ and $n_2$ and both are dependent on $x$).
Of course, that we couldn't find such $n_0$ is not a proof that the convergence is not uniform. But in this case, you can prove that $f_n$ is continuous in $[1,2]$, but converges in that interval to
$$f(x)=\left\{\begin{matrix}0 & 1<x<2\\ 1 & x=1 \vee x=2\\\end{matrix}\right.,$$
which is not continuous. And we know that if a sequence of continuous functions converges uniformly, the limit is necessarily a continuous function.
If you wanted to show that uniform convergence fails in the open interval $(1,2)$, the previous argument is not valid (since the limit is $0$ always, which is a continuous function). You should proove that there is some $\varepsilon>0$ such that for every $n_0$ you choose there is $n\ge n_0$ and $x\in (1,2)$ such that $|f_n(x)-f(x)|\ge \varepsilon$.
Say $\varepsilon=1/2$, and take $n$ arbitrarily: is easy to show that
$$|f_n(x)-f(x)|=\left(\frac x2\right)^n+\left(\frac1x\right)^n \ge \frac 12$$
if you chose $x$ close enough to $1$ or $2$, since 
$$\lim_{x\to1}\left(\frac x2\right)^n+\left(\frac1x\right)^n=\frac1{2^n}+1> \frac12$$
and
$$\lim_{x\to2}\left(\frac x2\right)^n+\left(\frac1x\right)^n=1+\frac1{2^n}>\frac12.$$
However, you can have uniform convergence if you 'move away a little bit' from $1$ and $2$. So, for instance, convergence is uniform on $[1.3,1.4]$, or in $[1.01,1.99]$ or $(1.01,1.99)$.
