A limit concerning the integral of $x^n$ I pondered if the general integral of $x^n$ could be used with limits to prove that $$\int x^{-1}dx=\ln(x)+C$$
I started with $$\int x^ndx=\frac1{n+1}x^{n+1}+C$$
Then, I took the limit as $n$ approached $-1$.
$$\lim_{n\to-1}\frac1{n+1}x^{n+1}+C$$
From graphing, I found that it was true, and that $C$ was something that tended towards $\infty$ in $\ln(x)=\frac1{n+1}x^{n+1}+C$
I also found from graphing, that $C$ may be found as a function of $n$:
$$C=-\frac1{n+1}$$
Or,
$$\ln(x)=\lim_{n\to-1}\frac1{n+1}x^{n+1}-\frac1{n+1}$$
I was wondering if there were a more direct way of proving the limit than this.
 A: Shifting $n$,
$$ \lim_{n\to-1}\left(\frac1{n+1}x^{n+1}-\frac1{n+1}\right) = \lim_{n \to 0} \frac{x^n - 1}{n} = \lim_{h \to 0} \frac{x^h - x^0}{h} = \left. \frac{d}{dy} x^y \right|_{y = 0}
$$
The right side is well known to be $\log x$ and a commonly used trick in integration. To prove it,
$$ \left. \frac{d}{dy} x^y \right|_{y = 0} = \left. \frac{d}{dy} e^{y \log x} \right|_{y = 0} = \left. \log x e^{y \log x} \right|_{y = 0} = \log x
$$
A: Consider:
$$
\int\limits_1^x t^{n - 1}dt = \frac{x^{n} - 1}{n}
$$
We wish to take the limit as $n\rightarrow 0$:
$$
\lim_{n\rightarrow 0} \int\limits_1^x t^{n - 1}dt = \lim_{n\rightarrow 0} \frac{x^n - 1}{n}
$$
You can certainly use L'Hospital's rule but I suspect it's somewhat circular to do so.  It requires knowing that $\frac{d}{dt}e^t = e^t$.  For $x > 1$ it's certainly the case that $x^n = e^{\ln(x)n}$.  This then gives, using L'Hospital's rule:
\begin{align}
\lim_{n\rightarrow 0} \frac{e^{\ln(x)n} - 1}{n} =&\ \lim_{n\rightarrow 0}\frac{\ln(x)x^n}{1}\\
=&\ \ln(x)
\end{align}
