Let $(f_n)$ be a sequence of functions with $f_n : [a,\infty) \to \Bbb R$ defined by $f_n(x) = \frac{\sin nx}{1+nx}$, for all $x \in [a,\infty)$ and $n \in \Bbb N$. Show that $f_n$ converges pointwise to $f(x)=0$ on $[a,\infty)$.
attempt: Let $\varepsilon > 0$ and $x \in [a,\infty)$ be arbitrary. Choose $n_0 \in \Bbb N$ with $\frac{1}{n_0} < \frac{\varepsilon |x|}{1+\varepsilon}$ such that for all $n \in \Bbb N$ with $n\ge n_0$, we have \begin{align*} |f_n(x)-0| = \left|\frac{\sin nx}{1+nx}\right| &\le \frac{1}{|1+nx|} \\ &\le \frac{1}{n|x|-1} \\ &\le \frac{1}{n_0|x|-1} \\ &< \varepsilon. \end{align*} Hence, $f_n \to f$ pointwise on $[a,\infty). \qquad \qquad \Box$
Does this correct? I have a little bit confuse since the number $a$ given in the problem above is doesn't clear that it whether a negative or a non-negative number anyway.
The number $a$ must be positive; otherwise, the problem makes no sense.
Note that\begin{align}\frac1{n_0|x|-1}<\varepsilon&\iff n_0|x|>1+\frac1\varepsilon\\&\iff n_0>\frac{\varepsilon+1}{|x|\varepsilon}.\end{align}Other than that, what you did is correct.