What is _ x→1 x^ (n^2-1) -1 x^ (n+1) -1 equal to, where n > 1 is a...

The given limit is $\lim\limits_{x\to1}\dfrac{x^{(n^2-1)}-1}{x^{(n+1)}-1}$.

As $x \to 1$, the expression takes the indeterminate form $\dfrac{0}{0}$.

Applying L'Hospital's rule, we differentiate the numerator and the denominator with respect to $x$:

$\lim\limits_{x\to1}\dfrac{(n^2-1)x^{n^2-2}}{(n+1)x^n}$

Substituting $x=1$, we get:

$\dfrac{n^2-1}{n+1}$

$\Rightarrow \dfrac{(n-1)(n+1)}{n+1}$

$\Rightarrow n-1$

Answer: $n-1$