In my attempt to understand fractal dimensions, I tried calculating the Minkowski dimension of some basic non-fractal spaces and ran into a limit I can't seem to conquer. For $n>2$,
$\lim_{x\to 0^+}\frac{n \left((1-x)^n+x-1\right)}{(1-x) \left((1-x)^n-(-x)^n-1\right)}$
I evaluated the limit in Mathematica and got $n-1$, which is exactly what it should be (the boundary of the $n$-sphere is indeed $n-1$ dimensional), but I'd appreciate a more human solution.
Let $1 - x = t$ so that $t \to 1^{-}$ and then we have \begin{align} L &= n\lim_{t \to 1^{-}}\frac{t^{n} - t}{t^{n} - 1 - (t - 1)^{n}}\notag\\ &= n\lim_{t \to 1^{-}}t\cdot\frac{t^{n - 1} - 1}{t - 1}\cdot\frac{t - 1}{t^{n} - 1 - (t - 1)^{n}}\notag\\ &= n(n - 1)\lim_{t \to 1^{-}}\dfrac{1}{\dfrac{t^{n} - 1}{t - 1} - (t - 1)^{n - 1}}\notag\\ &= n(n - 1)\frac{1}{n}\notag\\ &= (n - 1)\notag \end{align} This works on the assumption that $n$ is an integer greater than $1$.
First factor a $(1-x)$ out of the numerator, so the expression becomes $$n\frac{(1-x)^{n-1}-1}{(1-x)^n-(-x)^n-1}$$
Then taylor/binomial expand top and bottom to first order in $x$ to get $$ n\frac{-(n-1)x}{-nx} = n-1.$$