What is $\lim\limits_{n \to 0} \frac{d}{dx} \frac{1}{n} x^n$? A limit that I find rather intriguing is $\lim\limits_{n \to 0} \frac{d}{dx} \frac{1}{n} x^n$. Following the usual rules for differentiation of polynomials, this would be  $\lim\limits_{n \to 0} \frac {nx^{n-1}}{n} = x^{-1}$. It seems unlikely that this is actually the limit because the derivative of $ln(x)$ is $1/x$. Is there a way to prove what this actually is?
 A: The function $x^t$ for arbitrary real $t$ is only defined for $x>0$ as
$$
x^t=\exp(t\log x)
$$
(natural logarithm). Then, by the chain rule,
$$
\frac{d}{dx}x^t=\exp(t\log x)\cdot\frac{t}{x}=t
\exp(t\log x-\log x)=\exp((t-1)\log x)=tx^{t-1}
$$
Thus your limit is indeed
$$
\lim_{t\to0}x^{t-1}=\lim_{t\to0}\exp((t-1)\log x)=x^{-1}
$$
A: Actually,
that is true.
$\begin{array}\\
\lim_{n \to 0}\dfrac{x^n-1}{n}
&=\lim_{n \to 0}\dfrac{e^{n\ln x}-1}{n}\\
&=\lim_{n \to 0}\dfrac{1+n\ln x+O(n^2)-1}{n}\\
&=\lim_{n \to 0}\ln x + O(n)\\
&=\ln x\\
\end{array}
$
A: There is no mistake in your reasoning. What you may find confusing is that you started from something that seems to have nothing to do with $\ln(x)$. Except in the process you took $n$ to tend to $0$ at some point. So to be fair let's see what happens when you try to do that before taking the derivative. You will see that it doesn't really make sense, it tends to infinity everywhere. Except look closely at the shape of the curve: it tends to look more and more like $\ln(x)$, escaping up to infinity!
Now the derivative doesn't care about vertical translations: if you replace $f_n(x)$ by $f_n(x)-f_n(1)$, it will not tend to infinity any more, it will have the same derivative, and it will tend to $\ln(x)$.
See this output by Wolfram Alpha:

