Prove $\lim_{x \to 1} \frac{x^b - 1}{x - 1} = b$ Prove that
$$
  \lim_{x \to 1} \frac{x^b - 1}{x - 1} = b
$$
(No L'Hospital's rule, or series)
I'm not sure how to go about this.
I have that $x^b = e^{b \ln(x)}$, which gives
$$
  \lim_{x \to 1} \frac{x^b - 1}{x - 1} = 
  \lim_{x \to 1} \frac{e^{b \ln(x)}- 1}{x - 1}
$$
But this doesn't (seem to) do much.
I also have that $1 - \frac{1}{x} < \ln(x) < x - 1$.

Another approach, as indicated by the proposer:
\begin{align}
x^{b} &= e^{b \, \ln(x)} = e^{b \, \ln(1 - (1-x))} = Exp\left[-b \, \sum_{k=1}^{\infty} \frac{(1-x)^{k}}{k} \right] \\
&= 1 - b \, \left((1-x) + \frac{(1-x)^2}{2} + \mathcal{O}((1-x)^{3}) \right) + \frac{(-b)^{2}}{2!} \, \left( (1-x)^{2} + \mathcal{O}((1-x)^{3}) \right) \\
& \hspace{10mm} + \frac{(-b)^{3}}{3!} \, \mathcal{O}((1-x)^{3}) \\
&= 1 + b \, (x-1) + \frac{b \, (b-1)}{2!} \, (x-1)^{2} + \mathcal{O}((1-x)^{3}) 
\end{align}
which leads to
\begin{align}
x^{b} - 1 &= b \, (x-1) + \frac{b \, (b-1)}{2!} \, (x-1)^{2} + \mathcal{O}((1-x)^{3}) \\
\frac{x^{b} - 1}{x-1} &= b + \frac{b(b-1)}{2} \, (x-1) + \mathcal{O}((1-x)^{2}) \\
\lim_{x \to 1} \, \frac{x^{b} - 1}{x-1} &= b.
\end{align}
