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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 $1 - \frac{1}{x} < \ln(x) < x - 1$.

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    $\begingroup$ There is a typo (because $\lim_{x \to 0} \frac{x^b - 1}{x - 1} = 1$). Probably the correct statement is $\lim_{x \to 1} \frac{x^b - 1}{x - 1} = b$. $\endgroup$ – Pedro Jul 5 '17 at 23:22
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    $\begingroup$ This can't be true for general b. Consider b=2, where this simplifies to x+1, and the limit is 1. $\endgroup$ – Andrew Jul 5 '17 at 23:22
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    $\begingroup$ Well if @JackD'Aurizio wants to leave this as it is , and is actually able to provide an interesting answer with the limit as 1, then I'll leave it and ask again for the limit(x \to 0) I guess $\endgroup$ – baxx Jul 5 '17 at 23:31
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    $\begingroup$ @baxx The limit at $x \to 0$ is $1\,$, not $b\,$. That was pointed out in the very first comment, and requires no special proof since $x=0$ is not a singularity. Just set $x=0$ in the expression, and you get $\displaystyle \frac{-1}{-1}=1\,$. $\endgroup$ – dxiv Jul 5 '17 at 23:34
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    $\begingroup$ @dxiv i see, it's a double typo then, as I typed the wrong thing, and there's a typo in the text :) (the text has $\lim_{x \to 0}$) $\endgroup$ – baxx Jul 5 '17 at 23:36
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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 $\lim_{x\to 0}\frac{e^x-1}{x}=\lim_{x\to 0}\frac{e^{bx}-1}{bx}=1.$

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    $\begingroup$ The last part is incomplete, and certainly difficult. You need some sort of uniform continuity for such an argument to be valid - in general you are not allowed to swap the order of taking limits ($x \to 1$ vs $\frac{p}{q} \to b$). $\endgroup$ – mathguy Jul 5 '17 at 23:29
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    $\begingroup$ Jack, Why was my initial comment deleted, which simply asked about this answer acquiring three upvotes within 38 seconds of this post? I presume you flagged it. For what reason? $\endgroup$ – Namaste Jul 5 '17 at 23:34
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    $\begingroup$ I am not convinced. How does the continuity of the exponential function prove that your $f$ is a continuous function? (Answer: it doesn't; it takes a more complex proof, using uniform continuity.) $\endgroup$ – mathguy Jul 5 '17 at 23:38
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    $\begingroup$ lol: what you added is the solution I already gave. Good job! $\endgroup$ – mathguy Jul 5 '17 at 23:41
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    $\begingroup$ @amWhy: I have nothing to do with your comments being removed or flagged. What's the point in asking how an answer got $x$ upvotes in time $T$? I guess that means some users upvoted it - sorry for being Monsieur De Lapalisse. $\endgroup$ – Jack D'Aurizio Jul 5 '17 at 23:54
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This is a rate of variation, and its limit is the derivative of the function at $x=1$: $$\frac{x^b-1}{x-1}=\frac{x^b-1^b}{x-1}\to b\, x^{b-1}\Big\vert_{x=1}=b.$$

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$$ \frac{x^b - 1}{x - 1} = \frac{e^{b\ln x} - 1}{b\ln x} \cdot \frac{b\ln x}{x-1}$$.

Are you allowed to use $\lim_{y \to 0} \dfrac{e^y - 1}y = 1$ and $\lim_{x\to 1} \dfrac{\ln x}{x-1}= 1$?

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  • $\begingroup$ The and part is unnecessary, since $\lim_{y\to 0}\frac{e^y-1}{y}$ and $\lim_{x\to 1}\frac{\log x}{x-1}$ are the same limit through straightforward manipulations ($x\to e^x$ is continuous and injective). $\endgroup$ – Jack D'Aurizio Jul 5 '17 at 23:45
  • $\begingroup$ @JackD'Aurizio - correct, but in most cases when one is allowed to assume one of those two limits, they are allowed to assume the other one as well (without explaining that one derives from the other). $\endgroup$ – mathguy Jul 5 '17 at 23:46
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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 $f'(1)=b$.

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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}

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