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.

Dumping my answer here to a different but similar question which was labeled as a duplicate and would no longer accept the answer that I had already typed.
$\lim\limits_{t\to1}\frac{5}{1-t^5}-\frac{2}{1-t^2}=$
$\lim\limits_{t\to1}\frac{5}{(1-t)(1+t+t^2+t^3+t^4)}-\frac{2}{(1-t)(1+t)}=$
$\lim\limits_{t\to1}\frac{5+5t-2-2t-2t^2-2t^3-2t^4}{(1-t)(1+t+t^2+t^3+t^4)(1+t)}$
Clearly the numerator evaluates to $0$ when $t=1$ so $t-1$ must be a factor.
$\lim\limits_{t\to1}\frac{5+5t-2-2t-2t^2-2t^3-2t^4}{(1-t)(1+t+t^2+t^3+t^4)(1+t)}=$
$\lim\limits_{t\to1}\frac{2t^4+2t^3+2t^2-3t-3}{(t-1)(1+t+t^2+t^3+t^4)(1+t)}=$
$\lim\limits_{t\to1}\frac{2t^3+4t^2+6t+3}{(1+t+t^2+t^3+t^4)(1+t)}=$
$\frac{2+4+6+3}{(1+1+1+1+1)(1+1)}=\frac{15}{(5)(2)}=\frac32$
(Long division was used to obtain $\frac{2t^4+2t^3+2t^2-3t-3}{t-1}=2t^3+4t^2+6t+3$ )
