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 $1 - \frac{1}{x} < \ln(x) < x - 1$.
 A: This is a rate of variation, and its limit is the derivative of the function at $x=1$:
$$\frac{x^b-1}{x-1}=\frac{x^b-1^b}{x-1}\to b\, x^{b-1}\Big\vert_{x=1}=b.$$
A: $$ \frac{x^b - 1}{x - 1} = \frac{e^{b\ln x} - 1}{b\ln x} \cdot \frac{b\ln x}{x-1}$$.
Are you allowed to use $\lim_{y \to 0} \dfrac{e^y - 1}y = 1$ and $\lim_{x\to 1} \dfrac{\ln x}{x-1}= 1$?
A: It suffices to show that 
$$
\frac{d}{dx}(x^b)=bx^{b-1}.
$$
To this end note that
$$
\frac{d}{dx}(x^b)=\frac{d}{dx}(e^{b\log x})
=e^{b\log x}\times\frac{b}{x}
=bx^{b-1}.
$$
This involves knowing the derivative of $\log(x)$ and the chain rule (the derivative of $\exp x$ by can be inferred since it is the inverse). If you define $\log$ using the integral definition, its derivative can be computed using the FTC. 
Let $f(x)=x^{b}$. Then your limit is $f'(1)=b$.
A: If $b$ is a positive natural number the claim is straightforward, since
$$ x^b-1 = (x-1)\cdot\underbrace{(x^{b-1}+x^{b-2}+\ldots+x+1)}_{b\text{ terms}}. $$
If $b\in\mathbb{Q}^+$, with $b=\frac{p}{q}$, the claim follows by the substitution $x=z^q$ and the previous result.
For $b\in\mathbb{R}^+$, the claim follows by the continuity and differentiability of the exponential function, since $x^b = e^{b\log x}$.

Alternative approach: from $\lim_{x\to 0}\frac{e^x-1}{x}=1$ we have that
$$ \lim_{x\to 0}\frac{e^{bx}-1}{e^{x}-1} = \lim_{x\to 0}\frac{e^{bx}-1}{bx}\cdot\frac{x}{e^x-1}\cdot b = b $$
and by the substitution $x\mapsto \log w$ it follows that
$$ \lim_{w\to 1}\frac{w^b-1}{w-1}=b.$$

Prequel of the previous approach. Since $g(x)=e^x=g'(x)$ is a positive, increasing and convex function, it follows that for any $x\neq 0$
$$\frac{e^x-1}{x}=\frac{1}{x}\int_{0}^{x}e^t\,dt = 1+O(x)$$
hence $\lim_{x\to 0}\frac{e^x-1}{x}=\lim_{x\to 0}\frac{e^{bx}-1}{bx}=1.$
A: 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}
