$\lim _{x \to a} \frac{x^n-a^n}{x-a}$ when $n$ is irrational. Question on the theorem :$$\lim _{x \to a} \frac{x^n-a^n}{x-a}=na^{n-1}$$
Is it true when $n$ is irrational ? is there a proof ?
Ex: 
$$\lim _{x \to a} \frac{x^{\sqrt{2}}-a^{\sqrt{2}}}{x-a}=\sqrt{2}a^{\sqrt{2}-1}$$

Just another way to do it.
Define $$x=a(1+y)\implies A=\frac{x^n-a^n}{x-a}=\frac{a^n(1+y)^n-a^n}{ay}=a^{n-1}\frac{(1+y)^n-1}{y}$$ Apply the generalized binomial theorem  or Taylor series around $y=0$ $$(1+y)^n=1+n y+\frac{n(n-1) }{2}  y^2+O\left(y^3\right)$$ which makes $$\frac{(1+y)^n-1}{y}=n+\frac{n(n-1)}{2}  y+O\left(y^2\right)$$ $$A=a^{n-1}\left( n+\frac{n(n-1)}{2}  y+O\left(y^2\right)\right)$$ and $y\to 0$; then the limit $n a^{n-1}$.
If you want to go further, replace $y=\frac{x-a}a$ to get $$A=n a^{n-1}+\frac{n(n-1)}{2}a^{n-2}(x-a)+O\left((x-a)^2\right)$$ which shows the limit and how it is approached.
