<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] $\lim_\limits{x\to1}\frac {23(1-x^{11})-11(1-x^{23})}{(1-x^{11})(1-x^{23})}$
$\lim_\limits{x\to1}\frac {12-23x^{11}+ 11x^{23}}{(1-x^{11})(1-x^{23})}$
Now we could apply L'Hopitals at this point, or we can use algebra.
Using algebra, numerator and denominator both divide by $(x-1)^2$
$1-x^{11} = (1-x)\sum_\limits{i=0}^{10} x^i\\ 
1-x^{23} = (1-x)\sum_\limits{i=0}^{22} x^i$ 
$11x^{23} -23x^{11}+ 12 = (x-1)(11 x^{22}\cdots 11x^{11} - 12x^{10}\cdots +12)\\ =(x-1)^2 (11x^{21} + 2\cdot11 x^{20} + 3\cdot11 x^{19}\cdots +12\cdot 11 x^{10} + 12\cdot 10 x^9\cdots +12\cdot 2x + 12 $
Evaluated at 1.
The denominator:
$\sum_\limits{i=0}^{10} x^i = 11, \sum_\limits{i=0}^{22} x^i = 23$
the numerator:
$11 \sum_\limits{i=1}^{11} i + 12 \sum_\limits{i=1}^{11} i = (23)(11)(12)/2$
and the ratio  $= 6$
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