<s>[INST] Evaluate without L'Hopital: $\lim_{x\to1}\left\{\frac{9}{x^9-1}-\frac{5}{x^5-1}\right\}$ Evaluate the following limit without using L'Hospital method 
$$\lim_{x\to1}\left\{\frac{9}{x^9-1}-\frac{5}{x^5-1}\right\}$$
My turn is
$$L=\lim_{x\to1}\left\{\frac{9(x^5-1)-5(x^9-1)}{x^{14}-x^9-x^5+1}\right\}$$
$$L=\lim_{x\to1}\left\{\frac{9(x^5-1)-5(x^9-1)}{(x^{14}-1)-(x^9-1)-(x^5-1)}\right\}$$
then divide the numerator and the denominator by $$x-1$$
But I got again $$\frac{0}{0}$$
 [/INST] Note that
$$ x^n - 1 = (x-1)(x^{n-1}+x^{n-2}+\cdots + x + 1). $$
Thus we have
\begin{align}
\frac{9}{x^9-1} - \frac{5}{x^5-1} &= \frac{1}{x-1}\left[\frac{9}{\sum_{i=1}^9 x^{i-1}} - \frac{5}{\sum_{i=1}^5 x^{i-1}} \right]  \\
&=\frac{1}{x-1}\left[\frac{9\sum_{i=1}^5 x^{i-1} - 5\sum_{i=1}^9 x^{i-1}}{(\sum_{i=1}^9 x^{i-1})(\sum_{i=1}^5 x^{i-1})}\right].
\end{align}
And note that
\begin{align}
9\sum_{i=1}^5 x^{i-1} - 5\sum_{i=1}^9 x^{i-1}
&=4\sum_{i=1}^5 x^{i-1} - 5x^5\sum_{i=1}^4 x^{i-1} \\
&=4\sum_{i=1}^4 x^{i-1}(1-x^5) + x^4(4-x-x^2-x^3-x^4) \\
&=4(1-x)\left(\sum_{i=1}^4 x^{i-1}\right)\left(\sum_{i=1}^5 x^{i-1}\right) \\
&\hspace{0.5cm} +x^4(1-x)(1+(1+x)+(1+x+x^2)+(1+x+x^2+x^3))
\end{align}
Therefore,
\begin{align}
\frac{9}{x^9-1} - \frac{5}{x^5-1} &= -\frac{4(\sum_{i=1}^4 x^{i-1})(\sum_{i=1}^5 x^{i-1}) + x^4(4+3x+2x^2+x^3)}{(\sum_{i=1}^9 x^{i-1})(\sum_{i=1}^5 x^{i-1})}
\end{align}
and by letting $x \to 1$, we obtain
$$
\lim_{x \to 1} \left[\frac{9}{x^9-1} - \frac{5}{x^5-1}\right]
= -2.
$$
</s>