Proving the surprising limit: $\lim\limits_{n \to 0} \frac{x^{n}-y^{n}}{n}$ $=$ log$\frac{x}{y}$

By using L'Hospital rule we have \begin{align*} \lim_{n\to 0}\frac{x^n-y^n}{n}&=\lim_{n\to 0}\frac{x^n\log x-y^n\log y}{1}\\ &=\log x-\log y\\&=\log\frac{x}{y} \end{align*}

See here: Can a limit of an integral be moved inside the integral?

In general, passing the limit under the integral is not valid. In this circumstance, however, it is valid, as the functions converge uniformly on a finite interval. Uniform convergence means the following: if we let $M_n = \max\limits_{y\le t\le x}{|t^{n-1}-t^{-1}|}$ be the maximum difference of $t^{n-1}$ from $t^{-1}$, then $\lim\limits_{n\rightarrow 0}{M_n} = 0$. You can check that uniform convergence does hold here, so by a standard (and fairly easy to prove) theorem of analysis, passing the limit under the integral is justified.

Let $t = x/y$ and notice that \begin{align*} \lim_{n \to 0} \frac{x^n - y^n}{n} &= \lim_{n \to 0} y^n \frac{t^n - 1}{n} \\ &= \left(\lim_{n \to 0} y^n\right) \left(\lim_{n \to 0} \frac{t^n - 1}{n}\right) \\ &= 1 \cdot \frac{d}{dn}\Big|_{n = 0} t^n \\ &= \ln t = \ln \frac x y \end{align*}

as desired.