How find this limit $\lim_{x\to 1}\frac{f(2001)-f(2002)}{f(2002)-f(2003)}$ Let $$f(m)=\dfrac{m+1}{\dfrac{m+m+1}{\dfrac{m}{1-x^{m}}-\dfrac{m+1}{1-x^{m+1}}+\dfrac{1}{2}}+\dfrac{m+1+m+2}{\dfrac{m+1}{1-x^{m+1}}-\dfrac{m+2}{1-x^{m+2}}+\dfrac{1}{2}}}$$
Find this limit
$$I=\lim_{x\to 1}\dfrac{f(2001)-f(2002)}{f(2002)-f(2003)}$$
My try: since Use (L'Hôpital's rule） we have
\begin{align*}&\lim_{x\to 1}\left(\dfrac{n}{1-x^n}-\dfrac{n+1}{1-x^{n+1}}\right)\\
&=\lim_{x\to1}\dfrac{n(1-x^{n+1})-(n+1)(1-x^n)}{(1-x^n)(1-x^{n+1})}\\
&=\lim_{x\to 1}\dfrac{n-nx^{n+1}-n-1+(n+1)x^n}{1-x^n-x^{n+1}+x^{2n+1}}\\
&=\lim_{x\to 1}\dfrac{-n(n+1)x^n+n(n+1)x^{n-1}}{-(n+1)x^n-nx^n+（2n+1)x^{2n}}\\
&=\lim_{x\to 1}\dfrac{-n^2(n+1)x^{n-1}+n(n-1)(n+1)x^{n-2}}{-n(n+1)x^{n-1}-n^2x^{n-1}+2n(2n+1)x^{2n-1}}\\
&=\dfrac{n(n+1)[n-1-n]}{-n^2-n-n^2+4n^2+2n}=-\dfrac{1}{2}
\end{align*}
This problem is creat a teacher of China, zhejiang university,it is said  that teacher want Floored the arrogance of the students.
then I can't,Thank you
 A: Let $x = 1-z$, we have
$$\begin{align}
\frac{m}{1-x^m} &= \frac{m}{1-(1-z)^m}\\
& = \frac{m}{1 - \left(1 - mz + \frac{m(m-1)}{2}z^2 - \frac{m(m-1)(m-2)}{6}z^3 + O(z^4)\right)}\\
& = \frac{1}{z - \frac{m-1}{2}z^2 + \frac{(m-1)(m-2)}{6}z^2 + O(z^3)}\\
& =\frac{1}{z}\left[ 1 - \left( -\frac{m-1}{2}z + \frac{(m-1)(m-2)}{6}z^2\right) + \left(\frac{m-1}{2} z\right)^2 + O(z^3)\right]\\
&= \frac{1}{z} + \frac{m-1}{2} + \frac{m^2-1}{12}z + O(z^2)
\end{align}
$$
So
$$\begin{align}
&\frac{m}{1-x^m} - \frac{m+1}{1-x^{m+1}} + \frac12\\
=& \left(\frac{1}{z} + \frac{m-1}{2} + \frac{m^2-1}{12}z\right)
- \left(\frac{1}{z} + \frac{m}{2} + \frac{m^2-2m}{12}z\right) + \frac12 + O(z^2)\\
=&-\frac{2m+1}{12} z + O(z^2)
\end{align}
$$
This gives us 
$$f(m) = \frac{m+1}{\frac{2m+1}{-\frac{2m+1}{12}z + O(z^2)} + \frac{2m+3}{-\frac{2m+3}{12}z + O(z^2)}}
= -\frac{m+1}{24} z + O(z^2)
$$
and hence
$$\lim_{x\to 1}\frac{f(2001)-f(2002)}{f(2002)-f(2003)}
= \lim_{z\to 0}\frac{-\frac{2002}{24} z + \frac{2003}{24} z + O(z^2)
}{-\frac{2003}{24} z + \frac{2004}{24} z + O(z^2)} =
\lim_{z\to 0}\frac{\frac{z}{24}+O(z^2)}{\frac{z}{24} + O(z^2)} 
= 1$$
A: Just for fun, I'm adding this long computation which doesn't involve the Hospital's rule and explains your (...). 
$$
\lim_{x \to 1}\left(\frac{n}{1-x^n}-\frac{n+1}{1-x^{n+1}}\right) \\
= \lim_{x \to 1} \frac {\frac{n}{1 + x + \cdots + x^{n-1}} - \frac{n+1}{1+x+\cdots+x^n} }{1-x} \\
= \lim_{x \to 1} \frac 1{1+x+\cdots+x^{n-1}} \frac 1{1+x+\cdots +x^n} \frac{n(1 + x + \cdots + x^n) - (n+1)(1+x+\cdots+x^{n-1}) }{1-x} \\
= \frac 1{n(n+1)} \lim_{x \to 1} \frac {nx^n - (1+x+\cdots+x^{n-1})}{1-x} \\
= \frac{-1}{n(n+1)} \lim_{x \to 1} \frac{(x^n - 1) + (x^n - x) + \cdots + (x^n - x^{n-1})}{x-1} \\
= \frac{-1}{n(n+1)} \lim_{x \to 1} \frac{(x^n - 1) + x(x^{n-1} - 1) + \cdots + x^{n-1}(x - 1)}{x-1} \\
= \frac{-1}{n(n+1)} \lim_{x \to 1} \left( (1+x+\cdots+x^{n-1}) + x(1+x+\cdots + x^{n-2}) + \cdots + x^{n-2}(1+x) + x^{n-1}(1) \right) \\
= \frac{-1}{n(n+1)} \lim_{x \to 1} \sum_{i=0}^{n-1} \sum_{j=0}^{n-i} x^{i+j}\\ 
= \frac{-1}{n(n+1)} \left( n + (n-1) + \cdots + 1 \right) \\
= \frac {-1}2.
$$
