Showing that $\lim_{x \to 1} \left(\frac{23}{1-x^{23}}-\frac{11}{1-x^{11}} \right)=6$

Both fractions are unbounded as $x\rightarrow 1$. But if we rewrite $$\frac{23}{1-x^{23}}-\frac{11}{1-x^{11}}=\dfrac{23(1-x^{11})-11(1-x^{23})}{(1-x^{23})(1-x^{11})}=\dfrac{11x^{23}-23x^{11}+12}{1-x^{23}-x^{11}+x^{34}}$$ we can use L'Hopital since both sides tend to 0 as $x$ tends to 1. Differentiating both sides give $$\dfrac{253x^{22}-253x^{10}}{-23x^{22}-11x^{10}+34x^{33}}=253\dfrac{x^{12}-1}{-23x^{12}-11+34x^{23}}$$ Both sides still tend to 0, so we differentiate again and get $$253\dfrac{12x^{11}}{-276x^{11}+782x^{22}}$$ which tends to $$253\dfrac{12}{-276+782}=6$$

As $t\to0$, we have

$$\frac{23}{1-(1+t)^{23}}=-\frac{23}{23t+253t^2+O(t^3)}=-\frac{1}{t}\cdot\frac{1}{1+11t+O(t^2)}=-\frac{1}{t}\left(1-11t+O(t^2)\right)$$

Likewise

$$\frac{11}{1-(1+t)^{11}}=-\frac{11}{11t+55t^2+O(t^3)}=-\frac{1}{t}\cdot\frac{1}{1+5t+O(t^2)}=-\frac{1}{t}\left(1-5t+O(t^2)\right)$$

So the difference is

$$-\frac{1}{t}\left(1-11t-1+5t+O(t^2)\right)=6+O(t)$$

And your limit is $6$.

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.