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

Dividing numerator and denominator by $x-1$ should give $$\frac{9(x^4+x^3+x^2+x+1)-5(x^8+\cdots+x+1)}{(x^{13}+\cdots+1)-(x^8+\cdots+1)-(x^4+\cdots+1)}$$As you say, this gives $0/0$. We can write it as $$\require{cancel}\frac{-5x^8-5x^7-5x^6-5x^5+4x^4+4x^3+4x^2+4x+4}{x^{13}+x^{12}+x^{11}+x^{10}+x^9-x^4-x^3-x^2-x-1}\\=\frac{\cancel{(x-1)}(-5x^7-10x^6-15x^5-20x^4-16x^3-12x^2-8x-4)}{\cancel{(x-1)}(x^{12}+2x^{11}+3x^{10}+4x^9+5x^8+5x^7+5x^6+5x^5+5x^4+4x^3+3x^2+2x+1)}$$Cancel the $x-1$'s and take $x=1$. This gives: $$\frac{-90}{45}=-2$$
We chose to factor the $x-1$ since $1$ was a root of both the numerator and denominator, so $x-1$ must be a factor (and it was contributing the problematic $0$)
