Proving a sequence of functions converging pointwise doesn't converge uniformly on $(0,\infty)$ but does in $[a,\infty)$ where $a>0$ Let $f_n : (0,\infty) →\mathbb{R} , \,f_n(x) = \frac{1}{nx}$:
i) Prove that the $\lim_{n\rightarrow \infty} f_n(x)$ exists $\forall x > 0$.
ii )Prove that the convergence is not uniform in $(0,\infty)$ but it is in $[a,\infty)\, \forall a > 0. $
For the first step, I have attempted the result as follows:
$$\lim_{n\rightarrow \infty} \frac{1}{nx} = \frac{1}{x}\lim_{n\rightarrow \infty} \frac{1}{n} = \frac{1}{x}\frac{\lim_{n\rightarrow \infty}1}{\lim_{n\rightarrow \infty}n} = 0$$
Hence, we know that the function of $f_n \rightarrow f(x)=0$. I haven't been able to prove that it is not uniform in the $(0,\infty)$ interval, but it is in the $[a,\infty)$ interval.

Suppose $f_n$ converges to $f\equiv0$ uniformly on $(0,\infty)$. Then given $\varepsilon=\frac12$, there exists $N\in \mathbb N$ such that $|f_n(x)|\lt\frac12$ for $n\geq N$ and for all $x\in(0,\infty)$. In particular, this should hold for $x=\frac1N\in(0,\infty)$, a contradiction.
Now consider $[a,\infty)$ with $a\gt 0$. Given $\varepsilon\gt 0$, choose $N\in\mathbb N$ such that $N\gt\dfrac1{a\varepsilon}$. Then $$\left|\frac1{nx}\right|\leq \left|\frac1{na}\right|\lt \varepsilon$$ for all $n\geq N$ and for all $x\in [a,\infty) $.
