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

Michael Rozenberg
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baxx
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    There is a typo (because $\lim_{x \to 0} \frac{x^b - 1}{x - 1} = 1$). Probably the correct statement is $\lim_{x \to 1} \frac{x^b - 1}{x - 1} = b$. – Pedro Jul 05 '17 at 23:22
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    This can't be true for general b. Consider b=2, where this simplifies to x+1, and the limit is 1. – Andrew Jul 05 '17 at 23:22
  • I'm sorry - i've edited the comment (it was meant to be 0 not 1) – baxx Jul 05 '17 at 23:25
  • The limit is lim(x \to 0), sorry i put 1 there before, all fixed now – baxx Jul 05 '17 at 23:27
  • If the limit is $\lim_{x\to 0}$ then the limit is $1$, please perform a rollback, the limit equal to $b$ is the limit $\lim_{x\to \color{red}{1}}$. – Jack D'Aurizio Jul 05 '17 at 23:28
  • Could you let the asker do the editing, @Jack? Don't ask the OP to rollback and within 10 seconds, do it any way. – amWhy Jul 05 '17 at 23:30
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    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 – baxx Jul 05 '17 at 23:31
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    @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 $\displaystyle \frac{-1}{-1}=1,$. – dxiv Jul 05 '17 at 23:34
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    @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 $\lim_{x \to 0}$) – baxx Jul 05 '17 at 23:36
  • This is a pretty standard question and most notably the hard part is proving it in case $b$ is rational. One direct approach is provided in this answer https://math.stackexchange.com/a/1782225/72031 where I prove the general result $$\lim_{x\to a} \frac{x^{b} - a^{b}} {x-a} =ba^{b-1}$$ – Paramanand Singh Jul 06 '17 at 03:43

5 Answers5

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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 $\lim_{x\to 0}\frac{e^x-1}{x}=\lim_{x\to 0}\frac{e^{bx}-1}{bx}=1.$

Jack D'Aurizio
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  • I'm sorry but the original post had the wrong limit - it should have been lim(x \to 0) not lim(x \to 1) – baxx Jul 05 '17 at 23:28
  • @baxx: $$\lim_{x\to 0}\frac{x^b-1}{x-1}=1$$ is trivial, the interesting limit is the limit as $x\to 1$. – Jack D'Aurizio Jul 05 '17 at 23:29
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    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 $\frac{p}{q} \to b$). –  Jul 05 '17 at 23:29
  • @mathguy: We have a function defined on $\mathbb{Q}^+$ given by $f(b)=\lim_{x\to 1}\frac{x^b-1}{x-1}$ and for every $b\in\mathbb{Q}^+$ we have $f(b)=b$. The continuity of the exponential function allows to prove the continuity of $f$, hence the fact that $f(x)=x$ for any $x\in\mathbb{R}^+$. – Jack D'Aurizio Jul 05 '17 at 23:31
  • With this notation we have $$ f(b) = \lim_{z\to 0}\frac{e^{bz}-1}{e^z-1} = \lim_{z\to 0}\frac{e^{bz}-1}{z} $$ since $\lim_{z\to 0}\frac{e^z-1}{z}=1$. – Jack D'Aurizio Jul 05 '17 at 23:32
  • And once we have such well-known result $f(b)=b$ also follows from the substitution $z\mapsto w/b$. – Jack D'Aurizio Jul 05 '17 at 23:33
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    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? – amWhy Jul 05 '17 at 23:34
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    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.) –  Jul 05 '17 at 23:38
  • @mathguy: I added a second proof that is just as elementary. – Jack D'Aurizio Jul 05 '17 at 23:39
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    lol: what you added is the solution I already gave. Good job! –  Jul 05 '17 at 23:41
  • @mathguy: sorry, I did not notice it. It's a good way, indeed, so (+1) to you. – Jack D'Aurizio Jul 05 '17 at 23:44
  • For this part :

    $$ \lim_{x\to 0}\frac{e^{bx}-1}{e^{x}-1} = \lim_{x\to 0}\frac{e^{bx}-1}{bx}\cdot\frac{x}{e^x-1}\cdot b = b $$

    I don't follow why that must be true, given $\lim_{x\to 0}\frac{e^x-1}{x}=1$

     – baxx Jul 05 '17 at 23:54
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    @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. – Jack D'Aurizio Jul 05 '17 at 23:54
  • @baxx: if $\lim_{x\to 0}\frac{e^x-1}{x}=1$, then $\lim_{x\to 0}\frac{x}{e^x-1}=1$. The other limit follows from the substitution $x\mapsto bx$: $\lim_{x\to 0}\frac{e^{bx}-1}{bx}=1.$ – Jack D'Aurizio Jul 05 '17 at 23:55
  • The part where you deal with rational $b$ is not entirely trivial. It rather requires the continuity of $z=x^{1/q}$ as a function of $x$ (which follows via a standard but non-trivial theorem from the continuity of $x=z^{q} $ as a function of $z$). – Paramanand Singh Jul 06 '17 at 03:47
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This is a rate of variation, and its limit is the derivative of the function at $x=1$: $$\frac{x^b-1}{x-1}=\frac{x^b-1^b}{x-1}\to b\, x^{b-1}\Big\vert_{x=1}=b.$$

Bernard
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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 $\lim_{x\to 1} \dfrac{\ln x}{x-1}= 1$?

  • 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). – Jack D'Aurizio Jul 05 '17 at 23:45
  • @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). –  Jul 05 '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 $f'(1)=b$.

Sri-Amirthan Theivendran
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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}

Leucippus
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