<s>[INST] Showing that $\lim_{x \to 1} \left(\frac{23}{1-x^{23}}-\frac{11}{1-x^{11}} \right)=6$ How does one evaluate the following limit?
$$\lim_{x \to 1} \left(\frac{23}{1-x^{23}}-\frac{11}{1-x^{11}} \right)$$
The answer is $6$.
How does one justify this answer?
Edit: So it really was just combine the fraction and use L'hopital's rule twice (because function and its first derivative are of indeterminate form at $x=1$). This problem is more straightforward than it seems at first.
 [/INST] This answer does not use L'Hopital (personal taste), only a standard identity restated below, the binomial theorem, and a straightforward Taylor expansion to first order at $0$.
Using the identity $1-x^{2n+1} = (1-x)\sum_{k=0}^{2n} x^k$, we can rewrite
$$\begin{align*}
\frac{23}{1-x^{23}} -  \frac{11}{1-x^{11}} 
&= \frac{1}{1-x}\left(\frac{23}{\sum_{k=0}^{22}x^k} -  \frac{11}{\sum_{k=0}^{10}x^k} \right)\\
&= \frac{1}{1-x}\left(\frac{23\sum_{k=0}^{10}x^k}{\sum_{k=0}^{22}x^k\sum_{k=0}^{10}x^k} -  \frac{11\sum_{k=0}^{22}x^k}{\sum_{k=0}^{10}x^k\sum_{k=0}^{22}x^k} \right)\\
&= \frac{1}{\sum_{k=0}^{10}x^k\sum_{k=0}^{22}x^k}\cdot\frac{1}{1-x}\left(23\sum_{k=0}^{10}x^k - 11\sum_{k=0}^{22}x^k \right)\\
\end{align*}$$
Let us focus on the parenthesis (the first factor converges to $\frac{1}{11\cdot 23}$ by continuity, the second is the problematic one that will be "offset" by the parenthesis).
Writing $x=1+h)$ (where we will have $h\to 0$), we get, for any fixed integer $n$,
$$\begin{align*}
\sum_{k=0}^{n}x^k
&= \sum_{k=1}^{n}(1+h)^k
= \sum_{k=1}^{n} \sum_{\ell=0}^k \binom{k}{\ell} h^\ell \\
&= \sum_{k=0}^{n}(1+kh +o(h)) \\
&= (n+1)+\frac{n(n+1)}{2}h +o(h)
\end{align*}$$
when $h\to 0$, as $n$ is a constant. In particular, this implies
$$\begin{align*}
23\sum_{k=0}^{10}x^k - 11\sum_{k=0}^{22}x^k
&= 23\cdot 11+23\cdot \frac{11\cdot10}{2}h - 11\cdot 23-11\cdot \frac{22\cdot 23}{2}h + o(h)\\
&= 23\cdot 11\cdot (-6h) + o(h)\\
&= 23\cdot 11\cdot 6(1-x) + o(1-x)
\end{align*}$$
Overall, we thus have
$$\begin{align*}
\frac{23}{1-x^{23}} -  \frac{11}{1-x^{11}} 
&= \frac{23\cdot 11}{\sum_{k=0}^{10}x^k\sum_{k=0}^{22}x^k}\cdot\frac{6(1-x)+o(1-x)}{1-x} \\
&= \frac{23\cdot 11}{\sum_{k=0}^{10}x^k\sum_{k=0}^{22}x^k}\cdot (6+o(1)) \xrightarrow[x\to1]{} \frac{23\cdot 11}{23\cdot 11} \cdot 6 = 6
\end{align*}$$
as claimed.
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