Let $f_n(x) = \frac{1}{n^2x}$ and let $f(x)=0$. You may take it as a given that $f_n$ converges to $f$ pointwise for all $x\neq 0$.
(a) Prove that $f_n$ converges to $f$ uniformly on the interval $[\frac{1}{3}, 1]$.
(b) Prove that $f_n$ converges to $f$ non-uniformly on the interval $(0,1]$.
So, I'm a bit lost on how to prove both of these. Can someone kind of walk me through how to prove these using the definition of uniform convergence?
Hints:
Part a: Notice that whenever $x \geq 1/3$, we have that $\frac{1}{x} \leq 3$, so that $\frac{1}{n^2x} \leq \frac{3}{n^2}$. Then notice that $\frac{3}{n^2}$ converges to $0$ as $n \to \infty$.
Part b: Notice that $f_n(\frac{1}{n^2}) = 1$, which does not get close to $0$ when $n$ gets large.
Edit: More elaboration:
Part a: Let $\varepsilon>0$. Since $\frac{3}{n^2} \to 0$, there exists $N$ such that $\frac{3}{n^2}<\varepsilon$ whenever $n \geq N$. Thus, if $n \geq N$, we have that $|f_n(x)-f(x)| = \frac{1}{n^2x} \leq \frac{3}{n^2} < \varepsilon$, for all $x \in [\frac{1}{3},1]$. Therefore, by definition, $f_n \to 0$ uniformly on $[\frac{1}{3},1]$.
Part b: We want to show that $f_n$ does not converge uniformly to $0$ on $(0,1]$, so we need to show that there exists $\varepsilon >0$ such that for any $N$, there exists $n\geq N$ and $x \in (0,1]$ such that $|f_n(x)-f(x)|\geq \varepsilon$. Pick $\varepsilon = \frac{1}{2}$. Then for any $N \in \mathbb{N}$, set $n=N$ and $x = \frac{1}{n^2}$. Then $|f_n(x)-f(x)| = f_n(\frac{1}{n^2}) = 1 > \frac{1}{2} = \varepsilon$, as desired.