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 $$\frac{0}{0}$$

Observe that, by the standard binomial expansion,
$$
(1+h)^9-1=9h+36h^2+O(h^3),\qquad (1+h)^5-1=5h+10h^2+O(h^3)
$$ then, by setting $x=1+h$, as $x \to 1$ we have $h \to 0$, we get
$$
\begin{align}
\lim_{x\to1}\left(\frac{9}{x^9-1}-\frac{5}{x^5-1}\right)&=\lim_{h\to0}\left(\frac{9}{(1+h)^9-1}-\frac{5}{(1+h)^5-1}\right)
\\\\&=\lim_{h\to0}\left(\frac{9}{9h+36h^2+O(h^3)}-\frac{5}{5h+10h^2+O(h^3)}\right)
\\\\&=\lim_{h\to0}\left(\frac{1}{h+4h^2+O(h^3)}-\frac{1}{h+2h^2+O(h^3)}\right)
\\\\&=\lim_{h\to0}\frac{-2h^2+O(h^3)}{(h+4h^2+O(h^3))(h+2h^2+O(h^3))}
\\\\&=-2.
\end{align}
$$
