<s>[INST] 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!!?
 [/INST] You're on the right track. You can continue by dividing out the problematic factor $1-x$. What does long division of the numerator by $1-x$ give?
Hint:

 If $P(x)=(x-1)Q(x)$ then $P'(x)=Q(x)+(x-1)Q'(x)$, so $Q(1)=P'(1)$.

Details:

 Denote the polynomial in the numerator of the fraction by $P(x)$, so that$$P(x)=2017x^{2017}-x^{2016}-\ldots-x-1.$$By polynomial longdivision there exist polynomials $Q(x)$ and $R(x)$ with integer coefficients such that $P(x)=(x-1)Q(x)+R(x)$, where $\deg R(x)<\deg(x-1)$. That is to say $R(x)$ is constant, so $R(x)=R$ for some integer $R$. Plugging in $x=1$ shows that $$P(1)=(1-1)Q(1)+R(1)=R(1)=R,$$and it is not hard to see that $P(1)=0$. This means $P(x)=(x-1)Q(x)$ and so$$\lim\limits_{x\to 1}\dfrac{P(x)}{(1-x)(x^{2016} + \dots +1)(x^{2017} + \dots +1)}=\lim\limits_{x\to 1}\dfrac{Q(x)}{(x^{2016} + \dots +1)(x^{2017} + \dots +1)}.$$Here the hint comes in handy; we have\begin{eqnarray*}Q(1)&=&P'(1)=2017^2-2016-\ldots-2-1\\&=&2017^2-\frac12\cdot2017\cdot2016=2017\cdot1009.\end{eqnarray*}It follows that$$\lim\limits_{x\to 1}\dfrac{-Q(x)}{(x^{2016} + \dots +1)(x^{2017} + \dots +1)}=\frac{-2017\cdot1009}{2017\cdot2018}=-\frac12.$$

</s>