<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] Noting
$$ x^9-1=(x-1)\sum_{n=0}^8x^n, x^5-1=(x-1)\sum_{n=0}^4x^n $$
one has 
\begin{eqnarray}
&&\lim_{x\to1}\frac{9}{x^9-1}-\frac{5}{x^5-1}\\
&=&\lim_{x\to1}\frac{9\sum_{n=0}^4x^n-5\sum_{n=0}^8x^n}{(x-1)\sum_{n=0}^4x^n\sum_{n=0}^8x^n}\\
&=&\lim_{x\to1}\frac{9\sum_{n=0}^4(x^n-1)-5\sum_{n=0}^8(x^n-1)}{(x-1)\sum_{n=0}^4x^n\sum_{n=0}^8x^n}\\
&=&\lim_{x\to1}\frac{9\sum_{n=0}^4\sum_{k=0}^n(x-1)x^k-5\sum_{n=0}^8\sum_{k=0}^n(x-1)x^k}{(x-1)\sum_{n=0}^4x^n\sum_{n=0}^8x^n}\\
&=&\lim_{x\to1}\frac{9\sum_{n=0}^4\sum_{k=0}^nx^k-5\sum_{n=0}^8\sum_{k=0}^nx^k}{\sum_{n=0}^4x^n\sum_{n=0}^8x^n}\\
&=&\frac{9\sum_{n=0}^4(n+1)-5\sum_{n=0}^8(n+1)}{5\times9}\\
&=&-2.
\end{eqnarray}
</s>