Solve this limit $\lim\limits_{x\to 1}\left(\dfrac{2017}{1-x^{2017}} -\dfrac{2018}{1-x^{2018}}\right)$ \begin{align}
\lim\limits_{x\to 1}\left(\dfrac{2017}{1-x^{2017}} -\dfrac{2018}{1-x^{2018}}\right)
&= \lim\limits_{x\to 1}\dfrac{2017(x^{2017} + \dots + 1) - 2018(x^{2016} + \dots + 1)}{(1-x)(x^{2016} + \dots +1)(x^{2017} + \dots +1)}\\\\
&= \lim\limits_{x\to 1}\dfrac{2017x^{2017} - x^{2016} - \dots - 1}{(1-x)(x^{2016} + \dots +1)(x^{2017} + \dots +1)}
\end{align}
Did I do the right way with those step above?
I think next step is to separate:
$2017x^{2017} = \underbrace{x^{2017} + \dots + x^{2017}}_{\text{2017 addends}}$
Then combine with the rest and factorize $(1-x)$,
but it has a little confused after factorizing.
Please help me!!?

Yes, your approach is correct and you are doing fine. Now note that
\begin{align*}
2017x^{2017} - x^{2016} - \dots - 1&=2017(1+t)^{2017}-\frac{(1+t)^{2017}-1}{t}\\
&=2017(1+2017t+o(t))-\frac{1+2017t+\binom{2017}{2}t^2+o(t^2)-1}{t}\\
&=\left(2017^2-\frac{2017\cdot 2016}{2}\right)t+o(t)=(2017\cdot 1009)t+o(t)
\end{align*}
where $t=x-1\to 0$.
Hence, according to your work
$$\lim_{x\to 1}\left(\dfrac{2017}{1-x^{2017}} -\dfrac{2018}{1-x^{2018}}\right)
= \lim_{t\to 0}\dfrac{(2017\cdot 1009)t+o(t)}{-t(2017+o(1))(2018+o(1))}=-\frac{1}{2}.$$
Bonus exercise. More generally, show that for $n,m\in\mathbb{N}^+$,
$$\lim\limits_{x\to 1}\left(\dfrac{n}{1-x^{n}} -\dfrac{m}{1-x^{m}}\right)=\frac{n-m}{2}.$$
