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 $$\frac{0}{0}$$
 A: 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
$$
\begin{align}
\lim_{x\to1}\left(\frac{9}{x^9-1}-\frac{5}{x^5-1}\right)&=\lim_{h\to0}\left(\frac{9}{(1+h)^9-1}-\frac{5}{(1+h)^5-1}\right)
\\\\&=\lim_{h\to0}\left(\frac{9}{9h+36h^2+O(h^3)}-\frac{5}{5h+10h^2+O(h^3)}\right)
\\\\&=\lim_{h\to0}\left(\frac{1}{h+4h^2+O(h^3)}-\frac{1}{h+2h^2+O(h^3)}\right)
\\\\&=\lim_{h\to0}\frac{-2h^2+O(h^3)}{(h+4h^2+O(h^3))(h+2h^2+O(h^3))}
\\\\&=-2.
\end{align}
$$
A: 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 $0$)
A: 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
$$
\lim_{x \to 1} \left[\frac{9}{x^9-1} - \frac{5}{x^5-1}\right]
= -2.
$$
A: 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}
