<s>[INST] 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
 [/INST] Just for fun, I'm adding this long computation which doesn't involve the Hospital's rule and explains your (...). 
$$
\lim_{x \to 1}\left(\frac{n}{1-x^n}-\frac{n+1}{1-x^{n+1}}\right) \\
= \lim_{x \to 1} \frac {\frac{n}{1 + x + \cdots + x^{n-1}} - \frac{n+1}{1+x+\cdots+x^n} }{1-x} \\
= \lim_{x \to 1} \frac 1{1+x+\cdots+x^{n-1}} \frac 1{1+x+\cdots +x^n} \frac{n(1 + x + \cdots + x^n) - (n+1)(1+x+\cdots+x^{n-1}) }{1-x} \\
= \frac 1{n(n+1)} \lim_{x \to 1} \frac {nx^n - (1+x+\cdots+x^{n-1})}{1-x} \\
= \frac{-1}{n(n+1)} \lim_{x \to 1} \frac{(x^n - 1) + (x^n - x) + \cdots + (x^n - x^{n-1})}{x-1} \\
= \frac{-1}{n(n+1)} \lim_{x \to 1} \frac{(x^n - 1) + x(x^{n-1} - 1) + \cdots + x^{n-1}(x - 1)}{x-1} \\
= \frac{-1}{n(n+1)} \lim_{x \to 1} \left( (1+x+\cdots+x^{n-1}) + x(1+x+\cdots + x^{n-2}) + \cdots + x^{n-2}(1+x) + x^{n-1}(1) \right) \\
= \frac{-1}{n(n+1)} \lim_{x \to 1} \sum_{i=0}^{n-1} \sum_{j=0}^{n-i} x^{i+j}\\ 
= \frac{-1}{n(n+1)} \left( n + (n-1) + \cdots + 1 \right) \\
= \frac {-1}2.
$$
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